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:
- 2
- 1
- 6
- 4
The Correct Option is A
Solution and Explanation
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).
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Activity :
\(\therefore\) MD \(\times\) DN = \(\boxed{\phantom{SD}}\) \(\times\) DS \(\dots\) (Theorem of internal division of chords)
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