Hello, aspiring mathematicians! Welcome to a really interesting and elegant way to find the equation of a circle. We've already learned about the general equation of a circle, and the standard form `(x - h)^2 + (y - k)^2 = r^2` when we know the center `(h, k)` and radius `r`. But what if you're given different information? What if you're only given the endpoints of a diameter?

Don't worry, there's a beautiful method for that, and it's surprisingly intuitive once you understand the core idea! Let's dive in.

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Unlocking the Circle: The Diameter Form

Imagine you're an architect and you've drawn a blueprint for a circular patio. You know the exact coordinates of two points that lie directly opposite each other, forming the diameter of the patio. How would you then describe *all* the other points that make up the edge of this circular patio? This is exactly what we're going to figure out!

The "diameter form" of a circle's equation is a powerful tool when you are given the coordinates of the two endpoints of its diameter. Let's say these two endpoints are `A(x1, y1)` and `B(x2, y2)`. Our goal is to find an equation that any other point `P(x, y)` on the circle *must* satisfy.

The Magic Geometric Property: Angle in a Semicircle

Before we jump into coordinates, let's revisit a fundamental geometric property of circles that you might have encountered in earlier classes. It's the bedrock of this entire method!

Key Concept: The angle subtended by a diameter at any point on the circumference of the circle is always a right angle (90 degrees).

Think of it like this: if you have a straight line segment (your diameter) and you pick *any* point on the circle's boundary, and then draw lines from that point to the two ends of the diameter, the angle formed by these two lines will *always* be 90 degrees. This holds true no matter where you pick that point on the circle!

Analogy: Imagine you're standing on the circumference of a giant circular sports arena. If you look at one end of the diameter, and then turn your head to look at the other end of the diameter, the total angle your head turned would be 90 degrees, no matter where you stood on the circumference. This is a powerful visual!

Translating Geometry to Coordinate Geometry: The Power of Slopes

Now, how do we use this amazing geometric fact in the world of coordinate geometry? Well, in coordinate geometry, how do we express that two lines are perpendicular (form a 90-degree angle)?

You got it! We use their slopes!

Recall: If two non-vertical lines are perpendicular, the product of their slopes is -1. That is, `m1 * m2 = -1`.

This is the bridge between our geometric intuition and the algebraic equation we want to derive.

Let's put it all together step-by-step:

1. Define our points:
* Let the two given endpoints of the diameter be `A(x1, y1)` and `B(x2, y2)`.
* Let `P(x, y)` be *any* arbitrary point on the circumference of the circle. This `P(x, y)` is the generic point that will form our equation.

2. Form the lines:
* Draw a line segment connecting `P` to `A` (let's call it `PA`).
* Draw another line segment connecting `P` to `B` (let's call it `PB`).

3. Apply the geometric property:
* According to our key concept, the angle `∠APB` must be `90°`.
* This means the line segment `PA` is perpendicular to the line segment `PB`.

4. Calculate the slopes:
* The slope of line `PA`, let's call it `m_PA`:
`m_PA = (y - y1) / (x - x1)` (Remember: `(y2 - y1) / (x2 - x1)` for two points)
* The slope of line `PB`, let's call it `m_PB`:
`m_PB = (y - y2) / (x - x2)`

5. Apply the perpendicularity condition:
* Since `PA` is perpendicular to `PB`, the product of their slopes must be `-1`.
* `m_PA * m_PB = -1`
* `[(y - y1) / (x - x1)] * [(y - y2) / (x - x2)] = -1`

6. Simplify the equation:
* Multiply both sides by `(x - x1)(x - x2)` to clear the denominators:
`(y - y1)(y - y2) = - (x - x1)(x - x2)`
* Bring all terms to one side:
`(x - x1)(x - x2) + (y - y1)(y - y2) = 0`

And there you have it! This is the beautiful and compact diameter form of the equation of a circle.

The Diameter Form Equation of a Circle:
If `(x1, y1)` and `(x2, y2)` are the coordinates of the endpoints of a diameter of a circle, then the equation of the circle is:
`(x - x1)(x - x2) + (y - y1)(y - y2) = 0`

This equation elegantly captures the condition that any point `(x, y)` on the circle forms a right angle with the diameter's endpoints.

Why is this form so useful?

Sometimes, finding the center and radius can be a bit more work if you're only given the diameter endpoints. For example, to find the center, you'd need the midpoint formula: `Center = ((x1 + x2)/2, (y1 + y2)/2)`. To find the radius, you'd need the distance formula between the center and one endpoint (or half the distance between the two endpoints). The diameter form allows you to bypass these intermediate calculations and directly write down the equation, saving you time and effort!

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Let's Practice with an Example!

Example 1: Basic Application

Find the equation of a circle whose diameter has endpoints `A(1, 2)` and `B(5, 6)`.

Step-by-step Solution:

1. Identify the given endpoints:
Let `(x1, y1) = (1, 2)`
Let `(x2, y2) = (5, 6)`

2. Recall the diameter form equation:
`(x - x1)(x - x2) + (y - y1)(y - y2) = 0`

3. Substitute the coordinates into the equation:
`(x - 1)(x - 5) + (y - 2)(y - 6) = 0`

4. Expand and simplify:
* Expand `(x - 1)(x - 5)`:
`x * x - 5 * x - 1 * x + 1 * 5 = x^2 - 6x + 5`
* Expand `(y - 2)(y - 6)`:
`y * y - 6 * y - 2 * y + 2 * 6 = y^2 - 8y + 12`

5. Combine the expanded terms:
`(x^2 - 6x + 5) + (y^2 - 8y + 12) = 0`
`x^2 + y^2 - 6x - 8y + 5 + 12 = 0`
`x^2 + y^2 - 6x - 8y + 17 = 0`

So, the equation of the circle is `x^2 + y^2 - 6x - 8y + 17 = 0`.

This is the general form of the circle's equation. You can see how directly we arrived at it without needing to calculate the center or radius explicitly first!

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CBSE vs. JEE Focus:

For CBSE exams, understanding this derivation and being able to apply the formula directly is crucial. You might be asked to derive it or solve direct problems like the example above. It's a standard and important part of coordinate geometry.

For JEE Main & Advanced, this concept is fundamental and will be used as a tool in more complex problems. You won't usually be asked to just find the equation given two points, but this form might be the quickest way to find a circle's equation when those two points are identified as diameter endpoints within a larger problem (e.g., finding the intersection of a line and a circle that forms a diameter of another circle, or problems involving properties of triangles inscribed in a circle). Knowing this shortcut can save valuable time during competitive exams.

Pro Tip: Always remember the underlying geometric property – the angle in a semicircle is a right angle. This will help you recall the formula even if you forget the exact algebraic steps!

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By understanding the intuition behind this formula and its derivation, you're not just memorizing a formula; you're building a deeper understanding of how geometry and algebra beautifully intertwine. Keep practicing, and you'll master this in no time!

Welcome, aspiring engineers! Today, we're going to dive deep into a very elegant and incredibly useful method for finding the equation of a circle: when you're given the coordinates of the endpoints of its diameter. This method is a favorite in competitive exams like JEE, as it tests your understanding of fundamental geometric properties and their algebraic representation.

Let's begin our journey!

### The Fundamental Geometric Principle: Angle in a Semicircle

Before we jump into the algebra, let's recall a crucial geometric property of circles. Imagine a circle. Now, pick any two points on its circumference and connect them to form a diameter. Let these points be A and B. Now, pick *any other point* P on the circumference of the circle. If you connect P to A and P to B, what kind of angle do you think $angle APB$ will be?

Key Concept: The angle subtended by a diameter at any point on the circumference of the circle is always a right angle (90 degrees). This is a fundamental theorem from Euclidean geometry, often stated as "Angle in a semicircle is a right angle."

This simple yet powerful property is the cornerstone of deriving the equation of a circle when diameter endpoints are given. Why? Because it gives us a condition of perpendicularity, which we can easily translate into an algebraic equation using slopes.

### Derivation: Equation of a Circle from Diameter Endpoints

Let's say the endpoints of the diameter are given as $A(x_1, y_1)$ and $B(x_2, y_2)$.
Let $P(x, y)$ be any arbitrary point on the circumference of the circle.
According to our fundamental principle, the angle $angle APB$ must be $90^circ$.

What does it mean for two lines (or line segments) to be perpendicular? Their slopes must multiply to -1 (provided neither line is vertical).

1. Slope of AP ($m_{AP}$):
The slope of a line segment joining $(x_1, y_1)$ and $(x, y)$ is given by:
$m_{AP} = frac{y - y_1}{x - x_1}$

2. Slope of BP ($m_{BP}$):
The slope of a line segment joining $(x_2, y_2)$ and $(x, y)$ is given by:
$m_{BP} = frac{y - y_2}{x - x_2}$

3. Perpendicularity Condition:
Since $AP perp BP$, their slopes satisfy:
$m_{AP} cdot m_{BP} = -1$

Substituting the slope expressions:
$left(frac{y - y_1}{x - x_1}
ight) cdot left(frac{y - y_2}{x - x_2}
ight) = -1$

4. Simplifying the Equation:
Multiply both sides by $(x - x_1)(x - x_2)$:
$(y - y_1)(y - y_2) = -1 cdot (x - x_1)(x - x_2)$

Rearrange the terms to bring everything to one side:
$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$

This is the required equation of the circle!

This form is incredibly powerful and direct. It directly incorporates the coordinates of the diameter's endpoints.

### Understanding the Derived Equation

Let's break down the equation: $mathbf{(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0}$

* The terms $(x - x_1)$ and $(x - x_2)$ represent the horizontal distances from the current point P(x, y) to the vertical lines passing through the diameter endpoints.
* Similarly, $(y - y_1)$ and $(y - y_2)$ represent the vertical distances from P(x, y) to the horizontal lines passing through the diameter endpoints.
* When expanded, this equation will transform into the general form of a circle's equation: $x^2 + y^2 + 2gx + 2fy + c = 0$.
Let's quickly verify this:
$x^2 - x_2x - x_1x + x_1x_2 + y^2 - y_2y - y_1y + y_1y_2 = 0$
$x^2 + y^2 - (x_1+x_2)x - (y_1+y_2)y + x_1x_2 + y_1y_2 = 0$

Comparing this to $x^2 + y^2 + 2gx + 2fy + c = 0$:
$2g = -(x_1+x_2) implies g = -frac{x_1+x_2}{2}$
$2f = -(y_1+y_2) implies f = -frac{y_1+y_2}{2}$
$c = x_1x_2 + y_1y_2$

Notice that the center of the circle is $(-g, -f) = left(frac{x_1+x_2}{2}, frac{y_1+y_2}{2}
ight)$, which is exactly the midpoint of the diameter! This confirms our algebraic derivation aligns perfectly with our geometric understanding.

### Alternative Approach (Center-Radius Form)

While the diameter form is direct, you can always use the standard center-radius form of a circle's equation, $(x-h)^2 + (y-k)^2 = r^2$.

1. Find the Center (h, k): The center of the circle is the midpoint of the diameter.
$h = frac{x_1 + x_2}{2}$
$k = frac{y_1 + y_2}{2}$

2. Find the Radius (r): The radius is half the length of the diameter.
Length of diameter $D = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Radius $r = frac{1}{2} sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
So, $r^2 = frac{1}{4} left[ (x_2 - x_1)^2 + (y_2 - y_1)^2
ight]$

3. Substitute into Center-Radius Form:
$left(x - frac{x_1 + x_2}{2}
ight)^2 + left(y - frac{y_1 + y_2}{2}
ight)^2 = frac{1}{4} left[ (x_2 - x_1)^2 + (y_2 - y_1)^2
ight]$

While this method also works and gives the same result (try expanding both forms and you'll see!), it involves more steps and potentially more calculations (squares, square roots, fractions) compared to the direct diameter form. The diameter form is more elegant and efficient when given the diameter's endpoints.

### Illustrative Examples

Let's solidify our understanding with a few examples.

#### Example 1: Basic Application

Problem: Find the equation of the circle whose diameter has endpoints at $A(1, 2)$ and $B(5, 4)$.

Solution:
Here, $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (5, 4)$.
Using the diameter form: $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$
Substitute the coordinates:
$(x - 1)(x - 5) + (y - 2)(y - 4) = 0$

Now, expand and simplify:
$(x^2 - 5x - x + 5) + (y^2 - 4y - 2y + 8) = 0$
$x^2 - 6x + 5 + y^2 - 6y + 8 = 0$
$x^2 + y^2 - 6x - 6y + 13 = 0$

This is the equation of the circle.

#### Example 2: Involving Unknowns or Conditions

Problem: A circle passes through the origin $(0,0)$ and has a diameter whose one endpoint is $(4,0)$. If the center of the circle lies on the line $y=x$, find the equation of the circle.

Solution:
Let the endpoints of the diameter be $A(x_1, y_1)$ and $B(x_2, y_2)$.
We are given one endpoint, say $A(4, 0)$.
Let the other endpoint be $B(x_2, y_2)$.
The circle passes through the origin $O(0,0)$. Since $(0,0)$ is a point on the circumference, and $A(4,0)$ is an endpoint of the diameter, this implies that $B(x_2, y_2)$ and $O(0,0)$ might be the other endpoint of the diameter, or $A(4,0)$ and $O(0,0)$ could form a chord. Let's re-read carefully: "A circle passes through the origin $(0,0)$ and has a diameter whose one endpoint is $(4,0)$." This means the origin is *not necessarily* the other endpoint of the diameter. It's just a point on the circle.

Let the diameter endpoints be $(4,0)$ and $(x_2, y_2)$.
The center of the circle is $Cleft(frac{4+x_2}{2}, frac{0+y_2}{2}
ight) = Cleft(frac{4+x_2}{2}, frac{y_2}{2}
ight)$.
We are given that the center lies on the line $y=x$.
So, $frac{y_2}{2} = frac{4+x_2}{2} implies y_2 = 4+x_2$.

Now, the equation of the circle is $(x - 4)(x - x_2) + (y - 0)(y - y_2) = 0$.
$(x - 4)(x - x_2) + y(y - y_2) = 0$

Since the circle passes through the origin $(0,0)$, we can substitute $(x,y) = (0,0)$ into the equation:
$(0 - 4)(0 - x_2) + 0(0 - y_2) = 0$
$(-4)(-x_2) + 0 = 0$
$4x_2 = 0 implies x_2 = 0$.

Now that we have $x_2=0$, we can find $y_2$ using $y_2 = 4+x_2$:
$y_2 = 4+0 = 4$.

So, the endpoints of the diameter are $A(4, 0)$ and $B(0, 4)$.
Now, we can write the equation of the circle:
$(x - 4)(x - 0) + (y - 0)(y - 4) = 0$
$x(x - 4) + y(y - 4) = 0$
$x^2 - 4x + y^2 - 4y = 0$
$x^2 + y^2 - 4x - 4y = 0$

This problem combined the diameter form with conditions for the center and a point on the circle, making it a good JEE-level question.

#### Example 3: Finding the Other Endpoint and Equation

Problem: If one extremity of a diameter of the circle $x^2 + y^2 - 4x - 6y + 11 = 0$ is $(3, 2)$, find the coordinates of the other extremity and the equation of the circle using the diameter form.

Solution:
First, let's find the center of the given circle.
The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.
Comparing with $x^2 + y^2 - 4x - 6y + 11 = 0$:
$2g = -4 implies g = -2$
$2f = -6 implies f = -3$
The center of the circle is $(-g, -f) = (2, 3)$.

Let the given extremity of the diameter be $A(3, 2)$.
Let the other extremity be $B(x_2, y_2)$.
The center $(2, 3)$ must be the midpoint of the diameter $AB$.
Using the midpoint formula:
$frac{3 + x_2}{2} = 2 implies 3 + x_2 = 4 implies x_2 = 1$
$frac{2 + y_2}{2} = 3 implies 2 + y_2 = 6 implies y_2 = 4$

So, the other extremity of the diameter is $B(1, 4)$.

Now we have both endpoints of the diameter: $A(3, 2)$ and $B(1, 4)$.
Using the diameter form equation: $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$
$(x - 3)(x - 1) + (y - 2)(y - 4) = 0$

Expand and simplify:
$(x^2 - x - 3x + 3) + (y^2 - 4y - 2y + 8) = 0$
$x^2 - 4x + 3 + y^2 - 6y + 8 = 0$
$x^2 + y^2 - 4x - 6y + 11 = 0$

This is the original equation of the circle, which serves as a perfect check for our calculations! This type of question tests your ability to work backwards from the general form and then apply the diameter form.

### JEE Main & Advanced Perspective: Key Takeaways

* Efficiency: The diameter form $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ is the most efficient method when the endpoints of the diameter are directly given. Avoid the center-radius method in this specific scenario to save time and reduce calculation errors.
* Geometric Intuition: Always remember the underlying geometric principle – "Angle in a semicircle is a right angle." This intuition helps in visualizing the problem and understanding *why* the formula works.
* Disguised Problems: JEE problems might not explicitly state "endpoints of the diameter." Instead, they might provide conditions from which you can deduce the diameter's endpoints. For instance:
* A line segment is the diameter.
* A circle passes through two points, and its center lies on a perpendicular bisector of the segment connecting these points (if that segment is a chord).
* A circle passes through two points A and B, and $angle APB = 90^circ$ for a third point P on the circle, implying AB is a diameter.
* Verification: After finding the equation, you can always quickly verify it by finding the center (midpoint of diameter) and radius (half distance of diameter) and checking if they match. Also, substituting the given endpoints into the derived equation should make it true.
* Avoid Common Errors: Be careful with signs when substituting coordinates and expanding the terms. A single sign error can lead to a completely different equation.

### Conclusion

The equation of a circle with endpoints of diameter given is a beautiful example of how a simple geometric property translates into a concise algebraic formula. Mastering this concept is crucial for your coordinate geometry foundation. By understanding the derivation and practicing with various examples, you'll be well-prepared to tackle complex problems in JEE and other competitive examinations. Keep practicing, and visualize the geometry behind the algebra!

When the endpoints of a diameter of a circle are given as $A(x_1, y_1)$ and $B(x_2, y_2)$, remembering the formula for the circle's equation can be crucial for quick problem-solving in exams. This specific form is highly efficient as it bypasses the need to first calculate the center and radius.

Equation of a Circle with Endpoints of Diameter Given

The equation of a circle whose diameter has endpoints $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by:
$$ mathbf{(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0} $$

This formula is a direct and powerful tool. Let's explore mnemonics and conceptual shortcuts to ingrain this into your memory.

Mnemonics for Direct Recall

Here's a simple mnemonic to help you remember the structure of this equation:

* "X-Factors times X-Factors, PLUS Y-Factors times Y-Factors, EQUALS ZERO."

Let's break it down:

• X-Factors: Refers to the terms involving 'x' coordinates. You subtract each x-coordinate of the diameter's endpoints from 'x' and multiply them: $(x - x_1)(x - x_2)$.


• Y-Factors: Similarly, for 'y' coordinates: $(y - y_1)(y - y_2)$.


• PLUS: These two products are added together.


• EQUALS ZERO: The entire expression is equated to zero.

Think of it as two sets of 'parentheses products' (one for X, one for Y) that sum to zero.

Mnemonic Phrase: "X by X, Y by Y, Add and Get Zero!"

This emphasizes $(x-x_1)(x-x_2)$ and $(y-y_1)(y-y_2)$ are the building blocks.

Conceptual Shortcut: Geometric Derivation

The most elegant and robust "shortcut" for remembering this formula is to understand its geometric basis. This method is often faster and less prone to errors than rote memorization, especially under exam pressure.

The key geometric property is:

• Angle in a Semicircle: The angle subtended by a diameter at any point on the circumference of the circle is always a right angle (90 degrees).

Let $P(x, y)$ be any point on the circle. Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be the endpoints of the diameter.
Since $angle APB = 90^circ$, the line segment $AP$ is perpendicular to the line segment $BP$.

For two lines to be perpendicular (and neither is vertical/horizontal), the product of their slopes must be -1.

1. Slope of AP ($m_{AP}$): $m_{AP} = frac{y - y_1}{x - x_1}$


2. Slope of BP ($m_{BP}$): $m_{BP} = frac{y - y_2}{x - x_2}$

Applying the perpendicularity condition:
$m_{AP} cdot m_{BP} = -1$
$left(frac{y - y_1}{x - x_1}
ight) cdot left(frac{y - y_2}{x - x_2}
ight) = -1$

Now, cross-multiply:
$(y - y_1)(y - y_2) = -(x - x_1)(x - x_2)$

Rearrange the terms to get the standard diameter form:
$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$

This derivation is quick and reinforces your understanding, making it a powerful conceptual shortcut for both JEE Main and board exams.

JEE Main Focus	CBSE Board Focus
For JEE, speed and accuracy are key. Using the direct formula or its quick derivation saves time compared to finding the center and radius first.	Understanding the geometric derivation helps secure full marks for explanation-based questions. Applying the formula directly is also accepted.

By mastering both the direct mnemonic and the underlying geometric principle, you'll confidently tackle problems involving the equation of a circle given its diameter endpoints.

Quick Tips: Equation of a Circle with Endpoints of Diameter Given

This section provides crucial quick tips to efficiently handle problems involving finding the equation of a circle when the coordinates of the endpoints of its diameter are given. Mastering these tips will save time in both board exams and competitive tests like JEE Main.

1. The Direct Diameter Form Formula

The most efficient way to find the equation of a circle given the endpoints of its diameter, say $A(x_1, y_1)$ and $B(x_2, y_2)$, is using the diameter form:

• 
  Formula:
  
  

  $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$

  
  


• 
  Origin: This formula is derived from the property that the angle subtended by a diameter at any point on the circumference is 90 degrees. If $P(x, y)$ is any point on the circle, then the slope of $AP$ multiplied by the slope of $BP$ is -1.
  


• 
  JEE Advantage: This form is highly recommended for JEE Main as it directly gives the equation without needing to calculate the center and radius separately, thereby saving valuable time.
  

2. Alternative Method (Center-Radius Form)

If you forget the direct diameter form, you can always revert to the standard center-radius form, $ (x-h)^2 + (y-k)^2 = r^2 $, using the following steps:

• 
  Step 1: Find the Center (h, k): The center of the circle is the midpoint of the diameter.
  
  
  
  • $h = frac{x_1 + x_2}{2}$
  
  
  • $k = frac{y_1 + y_2}{2}$
  
  
  
  


• 
  Step 2: Find the Radius (r): The radius is half the length of the diameter. You can find the diameter length using the distance formula between $(x_1, y_1)$ and $(x_2, y_2)$, then divide by 2. Alternatively, find the distance between the center $(h,k)$ and one of the endpoints, say $(x_1, y_1)$.
  
  
  
  • $r = frac{1}{2} sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
  
  
  • Alternatively, $r = sqrt{(h - x_1)^2 + (k - y_1)^2}$
  
  
  
  


• 
  Step 3: Substitute into Standard Form: Once $h, k,$ and $r$ are known, substitute them into $(x - h)^2 + (y - k)^2 = r^2$.
  


• 
  CBSE Relevance: This method is perfectly acceptable for CBSE board exams and provides a good understanding of the circle's properties.
  

3. Common Pitfalls and Verification Tips

• 
  Sign Errors: Be extremely careful with signs when substituting coordinates into the formulas, especially in the diameter form $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$. A common mistake is using $(x+x_1)$ instead of $(x-x_1)$.
  


• 
  Squaring Errors: When using the distance formula for the radius, ensure proper squaring and summation.
  


• 
  Verification: After obtaining the equation of the circle, a quick check can be done by substituting the coordinates of both given endpoints into the derived equation. If both points satisfy the equation, your answer is likely correct. This is a very effective self-checking mechanism for both JEE and board exams.
  


• 
  Equation Expansion: The diameter form will typically result in the general form of the circle equation, $x^2 + y^2 + 2gx + 2fy + c = 0$, after expansion. Be prepared to expand and simplify.
  

Example Walkthrough (Mental Check)

Suppose the diameter endpoints are $A(1, 0)$ and $B(3, 0)$.

• 
  Using Diameter Form:
  $(x - 1)(x - 3) + (y - 0)(y - 0) = 0$
  $(x^2 - 4x + 3) + y^2 = 0$
  $x^2 + y^2 - 4x + 3 = 0$
  


• 
  Using Center-Radius Form:
  Center $C = left(frac{1+3}{2}, frac{0+0}{2}
  ight) = (2, 0)$
  Radius $r = sqrt{(2-1)^2 + (0-0)^2} = sqrt{1^2} = 1$
  Equation: $(x - 2)^2 + (y - 0)^2 = 1^2$
  $x^2 - 4x + 4 + y^2 = 1$
  $x^2 + y^2 - 4x + 3 = 0$
  


• 
  Both methods yield the same result, confirming the calculation.
  

By internalizing these quick tips, you can approach problems related to the equation of a circle with diameter endpoints with confidence and efficiency. Good luck!

While the topic of finding the equation of a circle given the endpoints of its diameter might seem abstract in a mathematics curriculum, its underlying principle has several practical applications across various fields. The ability to define a circle precisely from just two extreme points is invaluable in scenarios where the maximum extent of a circular shape or path is known.

Here are some real-world applications:

• 
  1. Architectural and Civil Engineering Design:
  
  
  
  • Scenario: Architects and engineers frequently design circular elements such as domes, archways, circular windows, or even entire circular plazas. Often, the design constraints provide the two extreme points of the desired circular feature.
  
  
  • Application: If the width or height of an archway is fixed, and these points represent the ends of its circular span (diameter), the equation helps in precisely determining the curve for construction. This is crucial for calculating material requirements, ensuring structural integrity, and accurate plotting of the design on site. For example, knowing the two ground-level points where a semicircular arch will begin and end allows engineers to define the entire curve of the arch.
  
  
  
  


• 
  2. Robotics and Automation:
  
  
  
  • Scenario: Programming robotic arms to perform tasks along a circular path or defining a robot's operational workspace.
  
  
  • Application: If a robot gripper needs to move in a precise circular arc, and the start and end points of this arc (which form a diameter for a specific circular movement) are known, the equation can be used to program the trajectory. This ensures accuracy in manufacturing processes, such as welding along a circular seam or drilling holes in a perfect circle, by defining the precise coordinates the robot arm must follow.
  
  
  
  


• 
  3. Navigation and Sensor Coverage:
  
  
  
  • Scenario: Defining circular safe zones, exclusion zones, or the coverage area of a sensor or communication device.
  
  
  • Application: In geographic information systems (GIS) or GPS applications, if a user wants to define a circular boundary (e.g., a "no-fly zone" or a "delivery zone") by marking two diametrically opposite points on a map, the equation can immediately generate the entire circular perimeter. Similarly, the effective circular range of a radar, Wi-Fi hotspot, or a sound sensor can be defined if its maximum detectable span (diameter) is known.
  
  
  
  


• 
  4. Computer Graphics and Game Development:
  
  
  
  • Scenario: Rendering circular objects, defining collision boundaries, or creating circular effects like explosions or light sources in virtual environments.
  
  
  • Application: Game developers often need to create circular elements. If a circular blast radius for an explosion or the effective area of a light source needs to be defined by its maximum two points of influence, the equation helps the game engine accurately draw and manage interactions within that circular region (e.g., "is player X affected by this circular blast?").
  
  
  
  

Understanding these applications helps to appreciate that seemingly theoretical mathematical concepts are fundamental tools for solving practical problems across diverse fields.

In mathematics, analogies help us connect abstract concepts to familiar real-world scenarios, making them more intuitive and memorable. For the equation of a circle when the endpoints of its diameter are given, the underlying geometric property is key.

### The Architectural Blueprint from a Structural Principle

Imagine you are an architect tasked with designing a perfectly circular arena. You are given only two pieces of critical information: the exact coordinates of two diametrically opposite points on the arena's boundary – say, the main entrance ($x_1, y_1$) and the emergency exit ($x_2, y_2$). Your goal is to define the entire circular boundary without explicitly being told the center or the radius.

Here's how this relates to the circle equation:

* The Diameter Endpoints ($x_1, y_1$) and ($x_2, y_2$): These are like your given, fixed reference points – the main entrance and emergency exit. They establish a crucial baseline for your design.
* The Fundamental Geometric Principle (Angle in a Semicircle is 90°): This is your core architectural rule. Any point 'P' on the arena's perimeter, when connected to both the main entrance and the emergency exit, *must* form a perfect 90-degree angle at point 'P'. This isn't just a suggestion; it's a defining structural characteristic for every point on the circular boundary relative to those two fixed points.
* The Equation $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$: This equation is your advanced CAD software or your precise architectural blueprint. It takes the coordinates of your two fixed reference points (entrance/exit) and mathematically translates the "90-degree angle rule" into a formula that defines *every single possible point* $(x, y)$ that lies on the circular perimeter. It bypasses the need to first find the center and radius, directly using the inherent geometric property to construct the shape.

In essence: Just as an architect uses fundamental structural principles (like how forces distribute or angles must meet) and critical reference points to define an entire structure, this equation uses the fixed endpoints of the diameter and the inherent 90-degree angle property to precisely define every point on the circle. It's a powerful and elegant way to conceptualize and derive the circle's equation.

For JEE Main aspirants, understanding this analogy helps in appreciating the derivation of the equation. It highlights that the equation is not just a formula to memorize, but a direct algebraic representation of a fundamental geometric property. This deeper understanding can be crucial for solving problems where the context might be slightly different but relies on the same core principle.

Prerequisites for Equation of a Circle with Endpoints of Diameter Given

Before diving into the equation of a circle when the endpoints of its diameter are given, it is crucial to have a strong foundation in several core concepts from Coordinate Geometry and basic Circle properties. Mastering these prerequisites will ensure a clear understanding and efficient problem-solving.

• 
  Basic Coordinate Geometry Concepts:
  
  
  
  • 
    Points and Coordinates: Understanding how to represent points $(x, y)$ in a Cartesian plane.
    
  
  
  • 
    Distance Formula: The ability to calculate the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ using the formula: $D = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. This is essential for finding the length of the diameter or the radius.
    
  
  
  • 
    Midpoint Formula: The ability to find the midpoint of a line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ using: $M = left(frac{x_1 + x_2}{2}, frac{y_1 + y_2}{2}
    ight)$. This is vital for finding the center of the circle, as the center is the midpoint of the diameter.
    
  
  
  
  


• 
  Standard Equation of a Circle:
  
  
  
  • 
    A thorough understanding of the general form of a circle's equation with center $(h, k)$ and radius $r$: $(x - h)^2 + (y - k)^2 = r^2$. Knowledge of this form is fundamental as the diameter form is derived from or leads back to this standard form.
    
  
  
  
  


• 
  Properties of a Circle:
  
  
  
  • 
    Definition of Diameter, Radius, and Center: Understanding that the diameter is a chord passing through the center, the radius is half the diameter, and the center is equidistant from all points on the circle.
    
  
  
  • 
    Angle in a Semicircle: A key geometric property stating that the angle subtended by a diameter at any point on the circumference is a right angle (90 degrees). This property forms the basis for the derivation of the diameter form of the circle's equation.
    
  
  
  
  


• 
  Slope of a Line and Perpendicular Lines:
  
  
  
  • 
    Knowing how to calculate the slope of a line segment connecting two points.
    
  
  
  • 
    Understanding the condition for perpendicular lines: if two non-vertical lines with slopes $m_1$ and $m_2$ are perpendicular, then $m_1 m_2 = -1$. This concept is indirectly used when applying the angle in a semicircle property to derive the equation.
    
  
  
  
  

JEE vs. CBSE Relevance: While these prerequisites are fundamental for both CBSE board exams and JEE Main, a deeper and more intuitive understanding, especially of the geometric properties like the angle in a semicircle, will be highly beneficial for JEE problem-solving, which often involves applying concepts in novel ways. Ensure you can apply these basic formulas quickly and accurately.

Understanding the equation of a circle when its diameter endpoints are given is a specific skill that builds upon several fundamental concepts in coordinate geometry. This section outlines the related topics that are either prerequisites, alternative methods, or direct extensions, crucial for a holistic understanding for both CBSE and JEE examinations.

Core Concepts & Prerequisites

Mastery of the following foundational topics is essential before tackling the diameter form of a circle's equation:

• Distance Formula:
  
  
  
  • JEE/CBSE Relevance: Absolutely fundamental. Used to calculate the length of the diameter, and consequently the radius of the circle. Knowing the distance between two points (x₁, y₁) and (x₂, y₂) as &sqrt;[(x₂ - x₁)² + (y₂ - y₁)²] is critical.
  
  
  
  


• Midpoint Formula:
  
  
  
  • JEE/CBSE Relevance: Directly applicable. The center of the circle is the midpoint of its diameter. The midpoint of (x₁, y₁) and (x₂, y₂) is given by ((x₁ + x₂)/2, (y₁ + y₂)/2). This helps in finding the center (h, k) for the standard form.
  
  
  
  


• Standard Equation of a Circle:
  
  
  
  • JEE/CBSE Relevance: The primary form to which the diameter method often leads. (x - h)² + (y - k)² = r² , where (h, k) is the center and r is the radius. This topic helps in understanding what you are ultimately trying to achieve.
  
  
  
  


• General Equation of a Circle:
  
  
  
  • JEE/CBSE Relevance: x² + y² + 2gx + 2fy + c = 0 . Understanding the conversion between standard and general forms, and how to extract the center and radius from the general form, is important for various problems. The diameter form (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0 is directly a variant of the general form.
  
  
  
  

Key Geometric Principle

• Angle Subtended by a Diameter at the Circumference:
  
  
  
  • JEE/CBSE Relevance: Crucial for understanding the derivation of the diameter form (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0. Any point (x, y) on the circle will form a right angle with the diameter endpoints (x₁, y₁) and (x₂, y₂). This property is expressed using slopes: (y - y₁)/(x - x₁) * (y - y₂)/(x - x₂) = -1.
  
  
  
  

Extended & Related Ideas

These topics provide a broader context and connect this concept to more complex problems:

• Locus of a Point:
  
  
  
  • JEE Relevance: Finding the equation of a circle with given diameter endpoints can be viewed as a locus problem: finding all points (x,y) such that the angle formed by (x,y) with the two diameter endpoints is 90 degrees.
  
  
  
  


• Equation of a Circle Passing Through Three Non-collinear Points:
  
  
  
  • JEE/CBSE Relevance: A more general method to find a circle's equation. The diameter form is a specific case if two of the three points happen to be diameter endpoints. This often involves solving simultaneous equations derived from the general form.
  
  
  
  


• Family of Circles:
  
  
  
  • JEE Relevance: Understanding how to write the equation of a circle passing through the intersection of a line and a circle, or two circles. The concept of a circle's equation is fundamental to these advanced topics.
  
  
  
  

By mastering these related concepts, students can approach problems involving circles with a stronger conceptual foundation and greater problem-solving flexibility.

When finding the equation of a circle given the endpoints of its diameter, students often fall into specific traps. Recognizing these pitfalls is crucial for securing marks in both board and competitive examinations.

Here are the common exam traps related to this concept:

• 
  Misinterpreting the Given Points:
  
  
  
  • 
    JEE Main/CBSE Trap: The most significant error is assuming any two given points on a circle are necessarily the endpoints of its diameter. The formula $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ is *strictly* applicable only when $(x_1, y_1)$ and $(x_2, y_2)$ are the diametrically opposite points.
    
  
  
  • 
    If a problem states "A circle passes through points A and B," these points are simply on the circumference. You cannot use the diameter form unless explicitly stated that A and B are endpoints of a diameter. If not, you might need to use the general equation of a circle, $x^2 + y^2 + 2gx + 2fy + c = 0$, and form equations by substituting the points, which is a longer process.
    
  
  
  
  


• 
  Sign Errors During Substitution:
  
  
  
  • 
    Substituting negative coordinates into the diameter form $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ is a frequent source of error. For instance, if $x_1 = -2$, the term becomes $(x - (-2))$, which simplifies to $(x + 2)$. A common mistake is to write $(x - 2)$.
    
  
  
  • 
    Tip: Always use parentheses when substituting negative values to avoid sign errors, e.g., $(x - (-2))(x - 3) + (y - 1)(y - (-4)) = 0$.
    
  
  
  
  


• 
  Arithmetic and Expansion Errors:
  
  
  
  • 
    After correct substitution, expanding the terms like $(x - x_1)(x - x_2)$ and $(y - y_1)(y - y_2)$ is where many calculation mistakes occur. For example, $(x-2)(x+3) = x^2 + x - 6$, not $x^2 + x + 6$ or $x^2 - x - 6$.
    
  
  
  • 
    Tip: Double-check the multiplication of terms and the combination of like terms. This is a fundamental algebraic step that needs precision.
    
  
  
  
  


• 
  Forgetting the RHS is Zero:
  
  
  
  • 
    The equation of a circle, like any equation, must be set equal to something. The diameter form of the circle's equation is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = mathbf{0}$. Sometimes, in a hurry, students might forget to equate the expanded expression to zero, leaving it as an expression instead of an equation.
    
  
  
  
  


• 
  Confusing with General Form Derivation:
  
  
  
  • 
    Some students might revert to finding the center (midpoint of diameter) and radius (half the distance between endpoints) and then using the standard form $(x-h)^2 + (y-k)^2 = r^2$. While this method is perfectly valid, it involves more steps (midpoint formula, distance formula, squaring the radius, expanding the binomials) and thus offers more opportunities for arithmetic errors.
    
  
  
  • 
    Tip: For problems specifically stating "endpoints of diameter," the form $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ is the most direct and least error-prone method. Use it efficiently.
    
  
  
  
  

By being mindful of these common traps, you can significantly improve your accuracy and speed when solving problems related to the equation of a circle with given diameter endpoints.

When the endpoints of a diameter of a circle are given, say $A(x_1, y_1)$ and $B(x_2, y_2)$, there are several powerful approaches to find the equation of the circle. Choosing the most efficient method is crucial for competitive exams like JEE.

Method 1: The Diameter Form (Recommended for JEE)

This is the most direct and generally fastest method for JEE Advanced and Main. It leverages the geometric property that the angle subtended by a diameter at any point on the circumference is 90 degrees. If $P(x, y)$ is any point on the circle, then the slope of AP multiplied by the slope of BP is -1 (for non-vertical/horizontal diameters).

• 
  Formula: The equation of the circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by:
  
  
  $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$
  


• 
  Approach:
  
  
  
  1. Identify the coordinates of the two endpoints of the diameter, $(x_1, y_1)$ and $(x_2, y_2)$.
  
  
  2. Directly substitute these values into the diameter form equation.
  
  
  3. Expand and simplify the equation to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.
  
  
  
  


• 
  JEE Insight: This form is particularly useful as it avoids intermediate calculations of the center and radius, saving time. It's robust and works for all cases, including vertical or horizontal diameters.
  

Method 2: Using Center-Radius Form

This method is more fundamental and often taught in CBSE, relying on finding the center and radius first. It's a good alternative if you prefer a two-step approach or if the question explicitly asks for the center or radius.

• 
  Steps:
  
  
  
  1. 
     Find the Center (h, k): The center of the circle is the midpoint of the diameter.
     
     
     $h = frac{x_1 + x_2}{2}$ and $k = frac{y_1 + y_2}{2}$
     
  
  
  2. 
     Find the Radius (r): The radius is half the length of the diameter. You can find the diameter length using the distance formula between $(x_1, y_1)$ and $(x_2, y_2)$, and then divide by 2. Alternatively, find the distance between the center $(h, k)$ and one of the endpoints, say $(x_1, y_1)$.
     
     
     $r = frac{1}{2} sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
     
     
     OR $r = sqrt{(h - x_1)^2 + (k - y_1)^2}$
     
  
  
  3. 
     Form the Equation: Substitute the center $(h, k)$ and radius $r$ into the standard equation of a circle:
     
     
     $(x - h)^2 + (y - k)^2 = r^2$
     
  
  
  4. Expand and simplify to the general form if required.
  
  
  
  


• 
  CBSE Relevance: This method builds on fundamental concepts of midpoint and distance formula, making it highly relevant for board exams and a strong foundation.
  

Example Problem:

Find the equation of the circle whose diameter has endpoints $(1, 2)$ and $(5, 8)$.

Method 1: Diameter Form

Method 2: Center-Radius Form

Given $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (5, 8)$. Using $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$: $(x - 1)(x - 5) + (y - 2)(y - 8) = 0$ $x^2 - 5x - x + 5 + y^2 - 8y - 2y + 16 = 0$ $x^2 + y^2 - 6x - 10y + 21 = 0$

Given $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (5, 8)$. 1. Center (h, k): $h = frac{1+5}{2} = 3$ $k = frac{2+8}{2} = 5$ Center $C(3, 5)$. 2. Radius (r): Distance from $C(3, 5)$ to $A(1, 2)$ $r = sqrt{(3-1)^2 + (5-2)^2} = sqrt{2^2 + 3^2} = sqrt{4 + 9} = sqrt{13}$ $r^2 = 13$ 3. Equation: $(x - h)^2 + (y - k)^2 = r^2$ $(x - 3)^2 + (y - 5)^2 = 13$ $x^2 - 6x + 9 + y^2 - 10y + 25 = 13$ $x^2 + y^2 - 6x - 10y + 21 = 0$

Both methods yield the same result, but the diameter form is clearly more direct.

Key Takeaways for Exam Success:

• For competitive exams, always prioritize the Diameter Form unless the center or radius is specifically asked for. It minimizes calculations and potential errors.


• Understand the geometric reasoning behind the diameter form (angle in a semicircle is 90 degrees) as it helps in related problems involving tangents or other properties.


• Practice converting the final equation to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$ as often required.

Mastering these approaches will ensure you can efficiently tackle problems involving the equation of a circle given the endpoints of its diameter.

This section highlights the key aspects of "Equation of a circle with endpoints of diameter given" that are particularly relevant for your CBSE board exams. While the fundamental concept is the same for JEE, CBSE often emphasizes different facets, especially the derivation and step-by-step solution presentation.

CBSE Focus Areas: Equation of a Circle with Diameter Endpoints

For CBSE board exams, understanding the derivation and the direct application of the diameter form of a circle's equation is crucial. Questions are generally straightforward, focusing on your ability to apply the formula correctly and perform algebraic manipulations accurately.

1. 
   Understanding the Diameter Form Equation:
   

   If the endpoints of a diameter of a circle are given as (x₁, y₁) and (x₂, y₂), the equation of the circle is:

   
   

   
   (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0
   

   
   

   This form is a direct and efficient way to write the equation when diameter endpoints are known.

   
   


2. 
   Derivation - A Key CBSE Skill:
   

   Unlike JEE, CBSE board exams might specifically ask for the derivation of this formula. This derivation is based on a fundamental geometric property:

   
   
   
   
   • Geometric Principle: The angle subtended by a diameter at any point on the circumference of a circle is always a right angle (90°).
   
   
   • Derivation Steps (using slopes):
     
     
     
     1. Let P(x, y) be any point on the circumference of the circle.
     
     
     2. Let A(x₁, y₁) and B(x₂, y₂) be the endpoints of the diameter.
     
     
     3. Since ∠APB = 90°, the line segment AP is perpendicular to BP.
     
     
     4. The product of the slopes of two perpendicular lines is -1 (provided neither line is vertical).
        
        
        
        • Slope of AP (m₁) = (y - y₁) / (x - x₁)
        
        
        • Slope of BP (m₂) = (y - y₂) / (x - x₂)
        
        
        
        
     
     
     5. Thus, m₁ * m₂ = -1.
     
     
     6. Substituting the slopes: [(y - y₁) / (x - x₁)] * [(y - y₂) / (x - x₂)] = -1.
     
     
     7. Rearranging gives: (y - y₁)(y - y₂) = -(x - x₁)(x - x₂).
     
     
     8. Finally, (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0.
     
     
     
     
   
   
   • Be prepared to explain or perform this derivation in an exam.
   
   
   
   


3. 
   Typical CBSE Question Patterns:
   
   
   
   • Direct Application: You will be given two points which are the endpoints of a diameter. You simply need to plug these coordinates into the formula and simplify the equation to its general form (x² + y² + 2gx + 2fy + c = 0).
   
   
   • Finding Center and Radius: After finding the equation in general form, you might be asked to identify the center (-g, -f) and radius √(g² + f² - c). This tests your knowledge of converting between forms.
   
   
   • Combined Problems: Sometimes, the endpoints of the diameter might be given indirectly, e.g., as the intersection points of a line and another curve, or as vertices of a rectangle whose diagonal is the diameter. However, for CBSE, these are typically simpler variations.
   
   
   
   


4. 
   CBSE vs. JEE Perspective:
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   
   

   Aspect	CBSE Board Exams	JEE Main
   Emphasis	Derivation, direct formula application, algebraic accuracy, step-by-step solutions.	Conceptual application, indirect problems, integration with other topics, speed, and accuracy in calculations.
   Question Type	"Find the equation of the circle...", "Derive the equation...", "Find center and radius..."	Multiple choice, often requiring multi-step reasoning, sometimes with conditions or properties of tangents/normals.

   
   


5. 
   Common Mistakes to Avoid in CBSE:
   
   
   
   • Algebraic Errors: Be meticulous with expansion of brackets and combining like terms. Sign errors are common.
   
   
   • Not Simplifying: Always simplify the equation to its general form (x² + y² + 2gx + 2fy + c = 0) unless otherwise specified.
   
   
   • Incorrect Formula Usage: Ensure you correctly substitute (x₁, y₁) and (x₂, y₂) into the formula.
   
   
   
   

Mastering this topic for CBSE means not just knowing the formula but understanding its origin and being able to apply it systematically with algebraic precision. Practice solving a variety of problems from your NCERT textbook and reference guides to solidify your understanding.

In JEE Main, understanding the equation of a circle when the endpoints of its diameter are given is a frequently tested concept, often integrated into more complex problems. This specific form provides a powerful and efficient way to define a circle geometrically.

The Diameter Form of a Circle's Equation

If the endpoints of a diameter of a circle are given as P(x₁, y₁) and Q(x₂, y₂), then the equation of the circle is given by:

(x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0

This equation is derived from the geometric property that the angle subtended by a diameter at any point on the circumference is 90 degrees. If R(x, y) is any point on the circumference, then PR is perpendicular to QR. The product of their slopes, mPR * mQR, must be -1. This leads directly to the diameter form.

JEE Focus Areas & Key Applications

• Direct Application: Be proficient in directly applying this formula when the diameter endpoints are explicitly given. It's often quicker than finding the center (midpoint of diameter) and radius (half the distance between endpoints) to use the standard form (x-h)² + (y-k)² = r².


• Geometric Interpretation (Perpendicularity): The underlying principle of perpendicularity is crucial.
  
  
  
  • This form inherently captures the condition that if a point (x, y) lies on the circle with diameter endpoints (x₁, y₁) and (x₂, y₂), then the line segment joining (x, y) to (x₁, y₁) is perpendicular to the line segment joining (x, y) to (x₂, y₂).
  
  
  • This property is fundamental for solving locus problems involving right angles.
  
  
  
  


• Circumcircle of a Right-Angled Triangle: If a triangle is right-angled, its hypotenuse is the diameter of its circumcircle. This formula becomes invaluable for finding the circumcircle's equation directly when the vertices of the right angle are given, and thus the hypotenuse endpoints.


• Finding Diameter Endpoints from Other Conditions: JEE problems often don't directly give the endpoints. You might need to find them first, for example, as:
  
  
  
  • Intersection points of a line and another curve.
  
  
  • Feet of perpendiculars from a point to lines.
  
  
  • Reflections of points.
  
  
  
  


• Family of Circles: This form can sometimes be seen as a specific case within the concept of a family of circles. For instance, the family of circles passing through the intersection of two lines L1 = 0 and L2 = 0 can be written as L1L2 + λC = 0. If L1 and L2 represent the lines from the diameter endpoints to a point on the circumference, their product forms part of the diameter equation.

CBSE vs. JEE Approach

• CBSE: Primarily focuses on direct application of the formula or finding the center and radius. Questions are generally straightforward.


• JEE: While direct questions exist, more often this concept is a stepping stone. You'll need to identify that a given situation implies diameter endpoints or a right angle, and then apply the formula efficiently. Speed and conceptual clarity are paramount.

Example Application

Problem: Find the equation of the circle passing through the origin (0,0) and whose diameter lies on the line joining the intersection points of the lines x+y=4 and x-y=2.

Solution:

1. First, find the intersection point of x+y=4 and x-y=2. Adding the equations gives 2x=6, so x=3. Substituting into x+y=4 gives 3+y=4, so y=1. Thus, one endpoint of the diameter is P(3,1).


2. The problem states the circle passes through the origin (0,0) and this point (0,0) lies on the circle. The problem statement implies that the line segment from (3,1) to (0,0) is *not* the diameter. It mentions "whose diameter lies on the line joining the intersection points of the lines x+y=4 and x-y=2." This means the intersection (3,1) is *one* endpoint of the diameter. The problem is subtly tricky.
   *A common interpretation mistake here would be to think (0,0) is the other diameter endpoint.*
   The phrase "whose diameter lies on the line joining the intersection points" is slightly ambiguous without another point or condition.
   Let's re-read: "diameter lies on the line joining the intersection points of the lines x+y=4 and x-y=2."
   This simplifies to: The diameter lies on the point (3,1). This implies (3,1) is the center or one of the endpoints. Given the form of the question, it's more likely that (0,0) is a point on the circle, and the diameter has one endpoint as (3,1) and the other endpoint such that the line through them passes through (3,1) and makes a right angle with the line connecting it to (0,0).
   
   Let's assume a slightly different but common type of question to fit the "endpoints of diameter" pattern better:
   Corrected Problem Interpretation: Find the equation of the circle whose diameter has endpoints as the origin (0,0) and the intersection point of lines x+y=4 and x-y=2.
   
   
   

   
   Corrected Solution:
   1. Find the intersection point of x+y=4 and x-y=2. This is P(3,1).
   2. The other endpoint of the diameter is the origin, O(0,0).
   3. Using the diameter form with (x1, y1) = (0,0) and (x2, y2) = (3,1):
   (x - 0)(x - 3) + (y - 0)(y - 1) = 0
   x(x - 3) + y(y - 1) = 0
   x² - 3x + y² - y = 0
   4. The equation of the circle is x² + y² - 3x - y = 0.
   

   
   

Mastering this form provides a powerful tool for quickly solving a variety of coordinate geometry problems in JEE.