The equation of the circle whose diameter is common chord to the circles \(x^{2}+y^{2}+2 a x+c=0\) and \(x^{2}+y^{2}+2 b y+c=0\) is

Let \(S_{1} \equiv x^{2}+y^{2}+2 a x+c=0\)
and \(S_{2} \equiv x^{2}+y^{2}+2 b y+c=0\)
Equation of common chord as a diameter of the third circle
\(
\begin{aligned}
S_{1}-S_{2} &=0 \\
a x-b y &=0 \\
y &=\frac{a x}{b}...(i)
\end{aligned}
\)
On putting the value of \(y\) in Eq. (i), we get
\(
\begin{array}{r}
x^{2}+\left(\frac{a^{2} x^{2}}{b^{2}}\right)+2 a x+c=0 \\
\left(a^{2}+b^{2}\right) x^{2}+2 a b^{2} x+c b^{2}=0
\end{array}
\)
Let \(x_{1}\) and \(x_{2}\) be the roots of the equation.
\(\therefore \quad x_{1}+x_{2}=\frac{-2 a b^{2}}{a^{2}+b^{2}}\) and \(x_{1} x_{2}=\frac{c b^{2}}{a^{2}+b^{2}}\)
Similarly, \(\left(a^{2}+b^{2}\right) y^{2}+2 b a^{2} y+c a^{2}=0\)
Let \(y_{1}\) and \(y_{2}\) be the roots of the equation \(\therefore \quad y_{1}+y_{2}=\frac{-2 a^{2} b}{a^{2}+b^{2}}\)
and \(\quad y_{1} y_{2}=\frac{c a^{2}}{a^{2}+b^{2}}\)
Now, the required equation of third circle \(\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=0\)
\(x^{2}-\left(x_{1}+x_{2}\right) x+x_{1} x_{2}+y^{2}-\left(y_{1}+y_{2}\right) y\)
\(+y_{1} y_{2}=0\)
\(x^{2}+y^{2}+\frac{2 a b^{2}}{a^{2}+b^{2}} \cdot x+\frac{2 b a^{2}}{a^{2}+b^{2}} \cdot y\)
\(+\frac{c b^{2}}{a^{2}+b^{2}}+\frac{c a^{2}}{a^{2}+b^{2}}=0\)
\(\Rightarrow x^{2}+y^{2}+\frac{2 a b^{2}}{a^{2}+b^{2}} \cdot x+\frac{2 a^{2} b}{a^{2}+b^{2}} \cdot y\)
\(+\frac{c\left(a^{2}+b^{2}\right)}{\left(a^{2}+b^{2}\right)}=0\)
\(\Rightarrow x^{2}+y^{2}+\frac{2 a b^{2}}{\left(a^{2}+b^{2}\right)} \cdot x+\frac{2 a^{2} b}{\left(a^{2}+b^{2}\right)} \cdot y+c=0\)