Let the circle and be a circle having centre at (–1, 0) and radius 2. If the line of the common chord of and intersects the y-axis at the point P, then the square of the distance of P from the centre of is: [2024]

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Solution

Q.
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: [2024]

1 2

2 6

3 4

4 1

Ans.
(1)

We have, $C_{1} : x^{2} + y^{2} - 2 (x + y) + 1 = 0$

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

For common chord, we have $C_{1} - C_{2} = 0$

$\Rightarrow x^{2} + y^{2} - 2 x - 2 y + 1 - x^{2} - y^{2} - 2 x + 3 = 0$

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

$\Rightarrow 2 x + y - 2 = 0$

Since, common chord intersects y-axis

So, x = 0

$\Rightarrow$ y = 2

So, point of intersection of common chord with y-axis is P(0, 2).

Required distance = ${(\sqrt{{(1 - 0)}^{2} + {(1 - 2)}^{2}})}^{2} = 2$ .

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