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

The Correct Option is A

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).