Question

Let $S_1$ be a circle passing through $A(0,1)$ and $B(-2,2)$ & $S_2$ is a circle of $radius=\sqrt{10}$ such that $AB$ is common chord of $S_1$ & $S_2$, find the equation of $S_2$.

Answer 1

Let, the equation of the circle $S_2$ be

$x^2+y^2+2gx+2fy+c=0$,

whose centre is $(-g,-f)$ and radius is $\sqrt{g^2+f^2-c}$.

According to the problem $AB$ is also a chord of $S_2$, so $S_2$ also passes through $A(0,1)$ and $B(-2,2)$.Then,

$1+2f+c=0$........(1) and

$4+4-4g+4f+c=0$........(2).

Subtracting (2) from (1) we get,

$7-4g+2f=0\Rightarrow 2(f+1)=4g-5$........(3).

Also, given radius of the circle $S_2$ is $\sqrt{10}$.

Then,

$\sqrt{g^2+(f+1{)}^2}=\sqrt{10}$......(4). [Since the distance of the centre to $A$ is the radius of $S_2$.]

Now, using the value of $(f+1)$ from (3) in (4) we get

$g^2+\frac{(4g-5{)}^2}{4}=$ $10$

or, $4g^2+16g^2-40g+25=40$

or, $20g^2-40g-15=0$

or, $4g^2-8g-3=0$

$\Rightarrow g=\frac{8\pm \sqrt{64+48}}{8}=\frac{2\pm \sqrt{7}}{2}$.

Using the value of $g$ in (3) we get,

$f=\frac{-3\pm 2\sqrt{7}}{2}$.

So, the equation of $S_2$ be

${(x+\frac{2\pm \sqrt{7}}{2})}^2+{(y+\frac{-3\pm 2\sqrt{7}}{2})}^2=10$.