Welcome, future engineers and mathematicians! Today, we're diving deep into a fundamental concept in coordinate geometry: the intersection of a line and a circle. Specifically, we'll focus on circles with their centre at the origin, which simplifies our equations but keeps the underlying principles robust for more complex scenarios.
Imagine a perfect circular pond and a straight path. How many times can the path touch or cross the pond? It could miss it entirely, brush against it at one point, or cut through it at two distinct points. Mathematically, these scenarios correspond to 'no intersection', 'tangency', and 'two distinct intersection points'. We'll explore two powerful methods to analyze these situations.
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I. The Algebraic Approach: Substitution Method
This method is straightforward: if a point lies on both the line and the circle, its coordinates must satisfy both equations simultaneously.
Let's begin with our standard equations:
1.
Equation of a circle with centre at the origin:
A circle with centre $(0,0)$ and radius $r$ has the equation:
$x^2 + y^2 = r^2$
2.
Equation of a straight line:
A general equation of a straight line can be written as:
$y = mx + c$ (slope-intercept form)
or
$Ax + By + C = 0$ (general form)
Step-by-step Derivation:
To find the points of intersection, we substitute the value of $y$ from the line equation into the circle equation.
Let's use $y = mx + c$:
Substitute $y = mx + c$ into $x^2 + y^2 = r^2$:
$x^2 + (mx + c)^2 = r^2$
Expand the term $(mx+c)^2$:
$x^2 + (m^2x^2 + 2mcx + c^2) = r^2$
Rearrange the terms to form a quadratic equation in $x$:
$(1+m^2)x^2 + (2mc)x + (c^2 - r^2) = 0$
This is a quadratic equation of the form $Ax^2 + Bx + C = 0$, where:
* $A = (1+m^2)$
* $B = (2mc)$
* $C = (c^2 - r^2)$
The nature of the roots of a quadratic equation is determined by its
discriminant ($Delta$), given by $Delta = B^2 - 4AC$.
Let's calculate the discriminant for our equation:
$Delta = (2mc)^2 - 4(1+m^2)(c^2 - r^2)$
$Delta = 4m^2c^2 - 4(c^2 - r^2 + m^2c^2 - m^2r^2)$
$Delta = 4m^2c^2 - 4c^2 + 4r^2 - 4m^2c^2 + 4m^2r^2$
$Delta = 4r^2 + 4m^2r^2 - 4c^2$
$Delta = 4(r^2(1+m^2) - c^2)$
Now, let's analyze the three possible cases based on the value of $Delta$:
Case 1: Two Distinct Intersection Points (Line is a Secant)
If $Delta > 0$, the quadratic equation has two distinct real roots for $x$. Each $x$ value will give a corresponding $y$ value from the line equation, resulting in two distinct points of intersection.
Condition: $4(r^2(1+m^2) - c^2) > 0 implies r^2(1+m^2) - c^2 > 0 implies c^2 < r^2(1+m^2)$.
Visually, the line cuts through the circle.
Case 2: One Intersection Point (Line is a Tangent)
If $Delta = 0$, the quadratic equation has two real and coincident roots for $x$. This means there is only one unique $x$ value, and thus one unique $y$ value, indicating that the line touches the circle at exactly one point.
Condition: $4(r^2(1+m^2) - c^2) = 0 implies r^2(1+m^2) - c^2 = 0 implies c^2 = r^2(1+m^2)$.
This is a crucial condition for tangency in JEE Mains & Advanced problems.
Visually, the line just grazes the circle.
Case 3: No Intersection Points
If $Delta < 0$, the quadratic equation has no real roots for $x$. This means there are no real coordinates $(x,y)$ that satisfy both equations, so the line does not intersect the circle.
Condition: $4(r^2(1+m^2) - c^2) < 0 implies r^2(1+m^2) - c^2 < 0 implies c^2 > r^2(1+m^2)$.
Visually, the line passes entirely outside the circle.
Example 1: Finding Two Distinct Intersection Points
Problem: Find the points of intersection of the line $x+y=5$ and the circle $x^2+y^2=13$.
Solution:
1. Rewrite the line equation in slope-intercept form: $y = 5-x$. Here, $m=-1$ and $c=5$.
2. The circle equation is $x^2+y^2=13$. So, $r^2=13$.
3. Substitute $y=5-x$ into the circle equation:
$x^2 + (5-x)^2 = 13$
$x^2 + (25 - 10x + x^2) = 13$
$2x^2 - 10x + 25 - 13 = 0$
$2x^2 - 10x + 12 = 0$
Divide by 2:
$x^2 - 5x + 6 = 0$
4. Solve this quadratic equation for $x$. We can factor it:
$(x-2)(x-3) = 0$
So, $x=2$ or $x=3$.
5. Find the corresponding $y$ values using $y=5-x$:
* If $x=2$, $y = 5-2 = 3$.
* If $x=3$, $y = 5-3 = 2$.
6. The points of intersection are
$(2,3)$ and $(3,2)$.
Example 2: Determining Tangency and Finding the Point of Tangency
Problem: Determine if the line $y = 2x+5$ is tangent to the circle $x^2+y^2=5$. If it is, find the point of tangency.
Solution:
1. Line: $y=2x+5 implies m=2, c=5$.
2. Circle: $x^2+y^2=5 implies r^2=5$.
3. Substitute $y=2x+5$ into $x^2+y^2=5$:
$x^2 + (2x+5)^2 = 5$
$x^2 + (4x^2 + 20x + 25) = 5$
$5x^2 + 20x + 25 - 5 = 0$
$5x^2 + 20x + 20 = 0$
Divide by 5:
$x^2 + 4x + 4 = 0$
4. This is a perfect square trinomial:
$(x+2)^2 = 0$
So, $x = -2$ (a repeated root).
5. Since we have a single, repeated root, the line is tangent to the circle.
6. Find the corresponding $y$ value using $y=2x+5$:
If $x=-2$, $y = 2(-2)+5 = -4+5 = 1$.
7. The point of tangency is
$(-2,1)$.
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II. The Geometric Approach: Perpendicular Distance Method
This method leverages the geometric relationship between the centre of the circle, its radius, and the line. It's often quicker to determine the *nature* of the intersection (how many points) than the algebraic method.
Let's use the general equation of a line:
$Ax + By + C = 0$.
The circle has its centre at $(0,0)$ and radius $r$, with equation $x^2+y^2=r^2$.
Step-by-step Derivation:
The distance 'd' from a point $(x_1, y_1)$ to a line $Ax+By+C=0$ is given by the formula:
$d = frac{|Ax_1 + By_1 + C|}{sqrt{A^2 + B^2}}$
For our case, the point is the centre of the circle $(0,0)$. So, $x_1=0$ and $y_1=0$.
The distance 'd' from the origin to the line $Ax+By+C=0$ is:
$d = frac{|A(0) + B(0) + C|}{sqrt{A^2 + B^2}} = frac{|C|}{sqrt{A^2 + B^2}}$
Now, we compare this distance 'd' with the radius 'r' of the circle:
Case 1: Two Distinct Intersection Points (Secant)
If the distance from the centre to the line ($d$) is less than the radius ($r$), the line must cut through the circle at two distinct points.
Condition: $d < r$
Case 2: One Intersection Point (Tangent)
If the distance from the centre to the line ($d$) is exactly equal to the radius ($r$), the line touches the circle at exactly one point (it's tangent).
Condition: $d = r$
Case 3: No Intersection Points
If the distance from the centre to the line ($d$) is greater than the radius ($r$), the line passes outside the circle and does not intersect it.
Condition: $d > r$
JEE Mains Focus: The geometric condition $d=r$ for tangency is incredibly powerful and often much faster than the algebraic discriminant method. You should be proficient in both.
Derivation of Tangency Condition using Geometric Method:
Let the line be $y=mx+c$. Convert it to general form: $mx - y + c = 0$.
Here, $A=m$, $B=-1$, $C=c$.
The distance from the origin $(0,0)$ to this line is $d = frac{|m(0) - 1(0) + c|}{sqrt{m^2 + (-1)^2}} = frac{|c|}{sqrt{m^2+1}}$.
For tangency, $d=r$:
$frac{|c|}{sqrt{m^2+1}} = r$
Square both sides:
$frac{c^2}{m^2+1} = r^2$
$c^2 = r^2(1+m^2)$
This is the same tangency condition derived using the discriminant method! This consistency reinforces our understanding.
Example 3: Determining the Nature of Intersection (Geometric Method)
Problem: Determine the nature of intersection of the line $3x+4y=15$ and the circle $x^2+y^2=9$.
Solution:
1. Circle: $x^2+y^2=9$. Centre $(0,0)$, radius $r = sqrt{9} = 3$.
2. Line: $3x+4y-15=0$. Here, $A=3, B=4, C=-15$.
3. Calculate the perpendicular distance 'd' from the origin to the line:
$d = frac{|3(0) + 4(0) - 15|}{sqrt{3^2 + 4^2}} = frac{|-15|}{sqrt{9+16}} = frac{15}{sqrt{25}} = frac{15}{5} = 3$.
4. Compare 'd' with 'r':
We found $d=3$ and $r=3$. Since $d=r$, the line is
tangent to the circle.
Example 4: Finding the Point of Tangency (Geometric Method)
Problem: Find the point of tangency for the line $3x+4y=15$ and the circle $x^2+y^2=9$ (from Example 3).
Solution:
For a tangent line, the point of tangency is the foot of the perpendicular drawn from the centre of the circle to the line.
1. Centre of the circle is $(0,0)$.
2. The given line is $3x+4y-15=0$. Its slope is $m_L = -A/B = -3/4$.
3. The line passing through the origin $(0,0)$ and perpendicular to $3x+4y-15=0$ will have a slope $m_P = -1/m_L = -1/(-3/4) = 4/3$.
4. Equation of the perpendicular line passing through $(0,0)$ is $y - 0 = frac{4}{3}(x - 0) implies y = frac{4}{3}x implies 4x - 3y = 0$.
5. The point of tangency is the intersection of the original line $3x+4y=15$ and the perpendicular line $4x-3y=0$.
We have a system of two linear equations:
(1) $3x + 4y = 15$
(2) $4x - 3y = 0$
From (2), $y = frac{4}{3}x$. Substitute this into (1):
$3x + 4left(frac{4}{3}x
ight) = 15$
$3x + frac{16}{3}x = 15$
Multiply by 3 to clear the denominator:
$9x + 16x = 45$
$25x = 45$
$x = frac{45}{25} = frac{9}{5}$
Now find $y$:
$y = frac{4}{3}x = frac{4}{3}left(frac{9}{5}
ight) = frac{12}{5}$
6. The point of tangency is
$left(frac{9}{5}, frac{12}{5}
ight)$.
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III. Advanced Concepts for JEE: Length of the Chord
When a line intersects a circle at two distinct points, the segment connecting these two points is called a
chord. We can find its length using geometry.
Let the circle be $x^2+y^2=r^2$ (centre $O(0,0)$, radius $r$).
Let the line be $Ax+By+C=0$, intersecting the circle at points $P$ and $Q$.
Let 'd' be the perpendicular distance from the centre $O$ to the line $PQ$. We know $d < r$ for a chord to exist.
Let $M$ be the midpoint of the chord $PQ$. Then $OM perp PQ$.
In the right-angled triangle $ riangle OMP$:
* Hypotenuse $OP = r$ (radius)
* Side $OM = d$ (perpendicular distance from origin to line)
* Side $MP = frac{PQ}{2}$ (half the length of the chord)
Using the Pythagorean theorem:
$OP^2 = OM^2 + MP^2$
$r^2 = d^2 + left(frac{PQ}{2}
ight)^2$
$left(frac{PQ}{2}
ight)^2 = r^2 - d^2$
$frac{PQ}{2} = sqrt{r^2 - d^2}$
So, the
length of the chord $PQ = 2sqrt{r^2 - d^2}$.
Example 5: Calculating the Length of a Chord
Problem: Find the length of the chord cut by the line $y=x+1$ on the circle $x^2+y^2=4$.
Solution:
1. Circle: $x^2+y^2=4$. Centre $(0,0)$, radius $r=sqrt{4}=2$.
2. Line: $y=x+1$, which can be written as $x-y+1=0$. Here, $A=1, B=-1, C=1$.
3. Calculate the perpendicular distance 'd' from the origin $(0,0)$ to the line $x-y+1=0$:
$d = frac{|1(0) - 1(0) + 1|}{sqrt{1^2 + (-1)^2}} = frac{|1|}{sqrt{1+1}} = frac{1}{sqrt{2}}$.
4. Now, use the chord length formula:
Length of chord $= 2sqrt{r^2 - d^2}$
$= 2sqrt{2^2 - left(frac{1}{sqrt{2}}
ight)^2}$
$= 2sqrt{4 - frac{1}{2}}$
$= 2sqrt{frac{8-1}{2}}$
$= 2sqrt{frac{7}{2}}$
$= 2 imes frac{sqrt{7}}{sqrt{2}}$
Rationalize the denominator:
$= frac{2sqrt{7}}{sqrt{2}} imes frac{sqrt{2}}{sqrt{2}} = frac{2sqrt{14}}{2} = sqrt{14}$.
5. The length of the chord is
$sqrt{14}$ units.
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IV. CBSE vs. JEE Focus
Aspect |
CBSE (Class XI/XII) |
JEE Mains & Advanced |
|---|
Core Understanding |
Understand the three cases of intersection (two points, one point, no points) using both algebraic (discriminant) and geometric (distance) methods. Ability to find intersection points and basic chord lengths. |
Deep conceptual understanding of both methods. Ability to quickly switch between them based on problem demand. Efficient use of the tangency condition $c^2 = r^2(1+m^2)$. |
Problem Complexity |
Direct application of formulas. Given line and circle equations, find intersection points or determine nature of intersection. |
JEE Mains: Problems involving unknown parameters (e.g., find 'k' for which a line is tangent). Combining line-circle intersection with other concepts like family of lines, locus, or geometric properties.
JEE Advanced: More abstract and generalized problems. Could involve finding range of parameters for specific intersection conditions, common tangents to two circles (extending the tangency idea), or optimization problems. |
Derivations |
Understanding the derivation of the quadratic equation from substitution. Intuitive understanding of the distance formula. |
Thorough understanding and ability to derive the tangency condition ($c^2=r^2(1+m^2)$) from both algebraic and geometric perspectives. Derivation of chord length. |
Speed & Accuracy |
Focus on correct steps for finding solutions. |
High emphasis on quick calculations and choosing the most efficient method to save time. Avoiding calculation errors under pressure. |
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V. Conclusion
You now have a robust toolkit to analyze the intersection of a line and a circle centered at the origin. Both the algebraic (substitution & discriminant) and geometric (perpendicular distance) methods are fundamental. Remember:
* The
algebraic method is universal for finding the actual intersection points and understanding the nature of roots.
* The
geometric method is often faster for determining *how many* intersection points there are and is particularly elegant for tangency conditions.
* The tangency condition
$c^2 = r^2(1+m^2)$ is a key result for quick problem-solving in competitive exams.
* The
length of the chord formula $2sqrt{r^2 - d^2}$ is a direct application of the geometric method.
Mastering these concepts will provide a strong foundation for tackling more complex problems involving circles and other conic sections in higher-level mathematics. Keep practicing with diverse problems to solidify your understanding!