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 :

Mathematics · Circles

JEE Main 2024 — 5 April, Shift 2 — Question 12

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 $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ intersects the y -axis at the point P , then the square of the distance of P from

the centre of $\mathrm{C}_{1}$ is :

Option A:
2
Correct
Option B:
1
Option C:
6
Option D:
4

Answer: A

Step-by-step solution

$\mathrm{S}_{1}: \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-2 \mathrm{y}+1=0$

$S_{2}: x^{2}+y^{2}+2 x-3=0$

Common chord $=\mathrm{S}_{1}-\mathrm{S}_{2}=0$

$-4 x-2 y+4=0$

$2 x+y=2 \Rightarrow P(0,2)$

$\mathrm{d}_{(\mathrm{c}, \mathrm{p})}^{2}=(1-0)^{2}+(2-1)^{2}=2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 2
Subject: Mathematics
Chapter: Circles
Topic: Equation of length of common chord

Answer: A

Step-by-step solution

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