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JEE Main 2024

05-04-2024 S2

Question

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 :

Select the correct option:

A

2

B

1

C

6

D

4

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} & S_1: x^2+y^2-2 x-2 y+1=0 \\ & S_2: x^2+y^2+2 x-3=0\end{aligned}$
Common chord $=\mathrm{S}_1-\mathrm{S}_2=0$
$ \begin{aligned} & -4 x-2 y+4=0 \\ & 2 x+y=2 \Rightarrow P(0,2) \\ & d_{(c, p)}^2=(1-0)^2+(2-1)^2=2 \end{aligned} $

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