The circle $x^{2} + y^{2} = 4$ cuts the circle $x^{2} + y^{2} + 2 x + 3 y - 5 = 0$ in A and B. Then the equation of the circle on AB as diameter, is

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$13 (x^{2} + y^{2}) - 4 x - 6 y - 50 = 0$

$x^{2} + y^{2} - 5 x + 2 y + 72 = 0$

none of these

$9 (x^{2} + y^{2}) + 8 x - 4 y + 25 = 0$

answer is A.

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Detailed Solution

The equation of the circle on AB as diameter is

$(x^{2} + y^{2} - 4) + λ (2 x + 3 y - 1) = 0 [U sing S_{1} + λ (S_{2} - S_{1}) = 0]$

$or, x^{2} + y^{2} + 2 λ x + 3 λ y - (λ + 4) = 0$

The equation of $A B$ is

$2 x + 3 y - 1 = 0$

Ii $A B$ is a diameter of (i), then centre lies on $A B$ .

$- 2 λ - \frac{9 λ}{2} - 1 = 0 \Rightarrow λ = - \frac{2}{13}$

$13 x^{2} + 13 y^{2} - 4 x - 6 y - 50 = 0$

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