Question

The equation of the circle which passes through the point of intersection of the circles $x^{2}+y^{2}-8 x-2 y+7=0$ and $x^{2}+y^{2}-4 x+10 y+8=0$ and having its centre on $y$-axis will be

• A. $x^{2}+y^{2}+22 x+9=0$
• B. $x^{2}+y^{2}+22 x-9=0$
• C. $x^{2}+y^{2}+22 y+9=0$
• D. $x^{2}+y^{2}+22 y-9=0$

Answer 1

Answer: (C)

Equation of circles passing through point of intersection of given two circles is
$x^{2}+y^{2}-8 x-2 y+7+\lambda\left(x^{2}+y^{2}-4 x+10 y+8\right)=0$ ... (i)
Centre of circle is $(4+2 \lambda, 1-5 \lambda)$.
Given that the centre lies on the $y$-axis then $4+2 \lambda=0$
$\Rightarrow \lambda=-2$
Hence, the required equation is
$x^{2}+y^{2}+22 y+9=0$.