Let the circle C₁: x²+y²-2(x+y)+1=0 and C₂ be a circle having centre at (-1,0) and radius 2 . If the line of the common chord of C₁ and C₂ intersects the y-axis at the point P, then the square of the distance of P from the centre of C₁ is:

Solution

Maths sir · 4 steps

Write both circles in general form and subtract: what is left is the common chord's line.

1. Both circles. C₁ is x²+y²-2x-2y+1=0, with centre (1,1) and radius 1. C₂ is (x+1)²+y²=4, which expands to x²+y²+2x-3=0.
2. Subtract them. With both x²+y² coefficients equal to 1, S₁ - S₂ = 0 is the line of the common chord. The centres are √5 apart, which lies between 2-1 and 2+1, so the circles really do cut each other.

   S₁ - S₂ = -4x - 2y + 4 = 0 ⇒ 2x + y = 2

3. Meet the y-axis. Put x = 0 to get y = 2, so P = (0, 2).
4. So. The squared distance from the centre (1,1) is (0-1)² + (2-1)² = 1 + 1 = 2, which is option (c).

The idea to keep. For two circles in general form with x²+y² coefficients both 1, S₁ - S₂ = 0 gives the common chord (the radical axis). Always expand both circles fully before you subtract.

Answer from the NTA final answer key. Solution by PrepWiser.