JEE Main 2024 (05 Apr Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Circles: Let the circle $C_{1} : x^{2} + y^{2} - 2 (x + y) + 1 = 0$ and $C_{2}$ be a circle having centre at $(- 1, 0)$ and radius 2. If the line of the common chord of $C_{1}$ and $C_{2}$ intersects the $y$ -axis at the point $P$ , then the square of the distance of $P$ from the centre of $C_{1}$ is :

Q: Why do we subtract the two circle equations? Does it always give the common chord?

Subtracting the equations $S_1 - S_2 = 0$ eliminates the quadratic terms $x^2$ and $y^2$, leaving you with a linear equation. This linear equation represents the radical axis. If the circles intersect, this line is exactly the common chord. It is a beautiful algebraic shortcut that saves us from finding the actual intersection points of the circles!

Q: Is there a specific order for $S_1 - S_2$? What if I do $S_2 - S_1$?

The order does not matter! Whether you calculate $S_1 - S_2 = 0$ or $S_2 - S_1 = 0$, you will arrive at the same linear equation. The line represents the same geometric locus regardless of the sign convention.

Q: Why did we set $x=0$ to find point $P$?

The problem states that $P$ is the intersection of the common chord with the $y$-axis. By definition, any point on the $y$-axis must have an $x$-coordinate of $0$. Setting $x=0$ is the standard way to find the $y$-intercept of any line.

Select Answer:

Visualized Solution

Analyzing Circle $C_{1}$

Given Circle $C_{1} : x^{2} + y^{2} - 2 (x + y) + 1 = 0$
Expanding the equation: $x^{2} + y^{2} - 2 x - 2 y + 1 = 0$
Center $C_{1} = (- g, - f) = (1, 1)$

Understanding Circle $C_{2}$

Center $= (- 1, 0)$
Radius $= 2$
Equation: $(x - (- 1))^{2} + (y - 0)^{2} = 2^{2}$

Expanding Equation of $C_{2}$

$(x + 1)^{2} + y^{2} = 4$
$x^{2} + 2 x + 1 + y^{2} = 4$
$x^{2} + y^{2} + 2 x - 3 = 0$

Concept of the Common Chord

The equation of the common chord is $S_{1} - S_{2} = 0$
$S_{1} : x^{2} + y^{2} - 2 x - 2 y + 1 = 0$
$S_{2} : x^{2} + y^{2} + 2 x - 3 = 0$

Setting up the Subtraction

$(x^{2} + y^{2} - 2 x - 2 y + 1) - (x^{2} + y^{2} + 2 x - 3) = 0$

Calculating the Common Chord

$- 2 x - 2 y + 1 - 2 x + 3 = 0$
$- 4 x - 2 y + 4 = 0$
Dividing by $- 2$ : $2 x + y - 2 = 0$

Finding Intersection Point $P$

Point $P$ is the intersection with the $y$ -axis.
Set $x = 0$ in $2 x + y - 2 = 0$

Coordinates of Point $P$

$2 (0) + y - 2 = 0 ⟹ y = 2$
Point $P = (0, 2)$

Distance to the Center of $C_{1}$

Distance $d$ between $P (0, 2)$ and $C_{1} (1, 1)$
We need to find $d^{2}$ .

Applying the Distance Formula

$d^{2} = (1 - 0)^{2} + (1 - 2)^{2}$
$d^{2} = (1)^{2} + (- 1)^{2}$

Final Calculation of $d^{2}$

$d^{2} = 1 + 1 = 2$
The square of the distance is 2.

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

We are given two circles,

C_{1}

and

C_{2}

. The first circle is defined by the equation:

x^{2} + y^{2} - 2 (x + y) + 1 = 0

Expanding this, we obtain:

x^{2} + y^{2} - 2 x - 2 y + 1 = 0

Comparing this to the general form

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

, we identify the center of

C_{1}

(1, 1)

Defining the Second Circle

The second circle,

C_{2}

, has a center at

(- 1, 0)

and a radius of

2

. Using the standard form

(x - h)^{2} + (y - k)^{2} = r^{2}

, we write:

(x + 1)^{2} + y^{2} = 2^{2}

Expanding and simplifying this expression:

x^{2} + 2 x + 1 + y^{2} = 4

x^{2} + y^{2} + 2 x - 3 = 0

The Elegance of the Common Chord

To find the common chord, we utilize the radical axis by subtracting the two circle equations,

S_{1} - S_{2} = 0

. This eliminates the quadratic terms and yields a linear equation:

(x^{2} + y^{2} - 2 x - 2 y + 1) - (x^{2} + y^{2} + 2 x - 3) = 0

Simplifying the terms, we get:

- 4 x - 2 y + 4 = 0

Dividing the entire equation by

- 2

, we arrive at the equation of the common chord:

2 x + y - 2 = 0

Finding Point P and the Final Distance

The common chord intersects the

y

-axis at point

P

. Since any point on the

y

-axis has an

x

-coordinate of

0

, we substitute

x = 0

into the line equation:

2 (0) + y - 2 = 0 \Rightarrow y = 2

Thus, the coordinates of point

P

are

(0, 2)

. We now calculate the square of the distance between

P (0, 2)

and the center of

C_{1}

, which is

(1, 1)

Using the distance formula

d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

d^{2} = (1 - 0)^{2} + (1 - 2)^{2}

d^{2} = 1^{2} + (- 1)^{2}

d^{2} = 1 + 1 = 2

Conclusion

The square of the distance between point

P

and the center of

C_{1}

is $2$ . By leveraging the radical axis, we bypassed the need to calculate complex intersection points, demonstrating the power of coordinate geometry in solving JEE-level problems efficiently.

Common Traps & FAQs

Why do we subtract the two circle equations? Does it always give the common chord?

Subtracting the equations

S_{1} - S_{2} = 0

eliminates the quadratic terms

x^{2}

and

y^{2}

, leaving you with a linear equation. This linear equation represents the radical axis. If the circles intersect, this line is exactly the common chord. It is a beautiful algebraic shortcut that saves us from finding the actual intersection points of the circles!

Wait, do I need to worry about whether the circles actually intersect?

Great observation! The radical axis exists even if the circles don't intersect. However, the problem explicitly mentions a 'common chord,' which implies the circles do intersect. In JEE problems, you can trust the premise, but it is always good practice to check the distance between centers versus the sum of radii if you have time.

Is there a specific order for

S_{1} - S_{2}

? What if I do

S_{2} - S_{1}

The order does not matter! Whether you calculate

S_{1} - S_{2} = 0

S_{2} - S_{1} = 0

, you will arrive at the same linear equation. The line represents the same geometric locus regardless of the sign convention.

Why did we set

x = 0

to find point

P

The problem states that

P

is the intersection of the common chord with the

y

-axis. By definition, any point on the

y

-axis must have an

x

-coordinate of

0

. Setting

x = 0

is the standard way to find the

y

-intercept of any line.

I got

d = 2

instead of

d^{2} = 2

. Did I miss something?

Read the question carefully! It asks for the 'square of the distance.' It is a common trap to stop at the distance

d

. Always re-read the final line of the question before bubbling your answer to ensure you are providing exactly what is requested.

Select Answer:

Visualized Solution

Analyzing Circle $C_{1}$

Understanding Circle $C_{2}$

Expanding Equation of $C_{2}$

Concept of the Common Chord

Setting up the Subtraction

Calculating the Common Chord

Finding Intersection Point $P$

Coordinates of Point $P$

Distance to the Center of $C_{1}$

Applying the Distance Formula

Final Calculation of $d^{2}$

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

Defining the Second Circle

The Elegance of the Common Chord

Finding Point P and the Final Distance

Conclusion

Similar Questions

Let $C$ be a circle with radius $10$ units and centre at the origin. Let the line $x + y = 2$ intersects the circle $C$ at the points $P$ and $Q$ . Let $M N$ be a chord of $C$ of length 2 unit and slope -1. Then, a distance (in units) between the chord $P Q$ and the chord $M N$ is

Let the locus of the mid points of the chords of circle $x^{2} + (y - 1)^{2} = 1$ drawn from the origin intersect the line $x + y = 1$ at $P$ and $Q$ . Then, the length of $P Q$ is :

If one of the diameters o of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord of another circle 'C' whose center is at (2, 1), then its radius is

Let a circle $C : (x - h)^{2} + (y - k)^{2} = r^{2}, k > 0$ , touch the x-axis at $(1, 0)$ . If the line $x + y = 0$ intersects the circle $C$ at $P$ and $Q$ such that the length of the chord $P Q$ is 2, then the value of $h + k + r$ is equal to ____.

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A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

Let a circle $C$ touch the lines $L_{1} : 4 x - 3 y + K_{1} = 0$ and $L_{2} : 4 x - 3 y + K_{2} = 0, K_{1}, K_{2} \in R$ . If a line passing through the centre of the circle $C$ intersects $L_{1}$ at $(- 1, 2)$ and $L_{2}$ at $(3, - 6)$ , then the equation of the circle $C$ is

Let $C_{1}$ and $C_{2}$ be the centres of the circles $x^{2} + y^{2} - 2 x - 2 y - 2 = 0$ and $x^{2} + y^{2} - 6 x - 6 y + 14 = 0$ respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $P C_{1} Q C_{2}$ is :

The circle passing through the point $(- 1, 0)$ and touching the y-axis at $(0, 2)$ also passes through the point

Select Answer:

Visualized Solution

Analyzing Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Understanding Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Expanding Equation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Concept of the Common Chord

Setting up the Subtraction

Calculating the Common Chord

Finding Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Distance to the Center of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Applying the Distance Formula

Final Calculation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

Defining the Second Circle

The Elegance of the Common Chord

Finding Point P and the Final Distance

Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the locus of the mid points of the chords of circle x2+(y−1)2=1 drawn from the origin intersect the line x+y=1 at P and Q. Then, the length of PQ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one of the diameters o of the circle x2+y2−2x−6y+6=0 is a chord of another circle 'C' whose center is at (2, 1), then its radius is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a circle C pass through the points (4,2) and (0, 2), and its centre lie on y+2=0 Then the length of the chord, of the circle C, whose mid-point is (1,2), is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one of the diameters of the circle x2+y2−2x−6y+6=0 is a chord to the circle with centre (2,1), then the radius of the circle is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A circle is given by x2+(y−1)2=1, another circle C touches it externally and also the x-axis, then the locus of its centre is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The circle passing through the point (−1,0) and touching the y-axis at (0,2) also passes through the point

Analyzing Circle $C_{1}$

Understanding Circle $C_{2}$

Expanding Equation of $C_{2}$

Finding Intersection Point $P$

Coordinates of Point $P$

Distance to the Center of $C_{1}$

Final Calculation of $d^{2}$

Let the locus of the mid points of the chords of circle $x^{2} + (y - 1)^{2} = 1$ drawn from the origin intersect the line $x + y = 1$ at $P$ and $Q$ . Then, the length of $P Q$ is :

If one of the diameters o of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord of another circle 'C' whose center is at (2, 1), then its radius is

Let a circle C pass through the points $(4, 2)$ and (0, 2), and its centre lie on $y + 2 = 0$ Then the length of the chord, of the circle C, whose mid-point is $(1, 2)$ , is:

If one of the diameters of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord to the circle with centre $(2, 1)$ , then the radius of the circle is

A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

The circle passing through the point $(- 1, 0)$ and touching the y-axis at $(0, 2)$ also passes through the point