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:

• A. 2
• B. 1
• C. 6
• D. 4

Answer 1

Answer: (A)

2

Answer 2

The equations of the circles are given as:

$S_1 : x^2 + y^2 - 2x - 2y + 1 = 0,$ $S_2 : x^2 + y^2 + 2x - 3 = 0.$

The equation of the common chord is obtained by subtracting $S_2$ from $S_1$:

$S_1 - S_2 = 0,$ $-4x - 2y + 4 = 0.$

Simplifying, we get:

$2x + y = 2 \quad \implies \quad y = 2 - 2x.$

Intersection with the y-axis To find the intersection point $P$ with the y-axis, set $x = 0$:

$y = 2 \quad \implies \quad P(0, 2).$

Distance Calculation Let $C_1, \text{centre} = (1, 1)$. The square of the distance between $P(0, 2)$ and the centre of $C_1$ is given by:

$d^2(C_1, P) = (1 - 0)^2 + (2 - 1)^2 = 1 + 1 = 2.$

Therefore, the correct answer is Option (1).