The circles \(x^2+y^2+2 \mathrm{a} x+\mathrm{c}=0\) and \(x^2+y^2+2 \mathrm{~b} y+\mathrm{c}=0\) touch each other externally, if

\(
\begin{aligned}
& x^2+y^2+2 \mathrm{a} x+\mathrm{c}=0 \\
& \Rightarrow x^2+2 \mathrm{a} x+\mathrm{a}^2+y^2=\mathrm{a}^2-\mathrm{c} \\
& \Rightarrow(x+\mathrm{a})^2+y^2=\left(\sqrt{\mathrm{a}^2-\mathrm{c}}\right)^2
\end{aligned}
\)
i.e., it is a circle with centre \((-a, 0)\) and radius
\(
\sqrt{\mathrm{a}^2-\mathrm{c}}
\)
Simiłarly,
\(
\begin{aligned}
& x^2+y^2+2 \mathrm{~b} y+\mathrm{c}=0 \\
& \Rightarrow x^2+(y+\mathrm{b})^2=\left(\sqrt{\mathrm{b}^2-\mathrm{c}}\right)^2
\end{aligned}
\)
i.e., it is a circle with centre \((0,-b)\) and
\(
\text { radius }=t \sqrt{b^2-c}
\)
\(\therefore \quad\) If circles touch externally, then we get Sum of radii \(=\) Distance between centres
\(
\begin{aligned}
& \Rightarrow \sqrt{a^2-c}+\sqrt{b^2-c}=\sqrt{a^2+b^2} \\
& \Rightarrow a^2-c+b^2-c+2 \sqrt{a^2-c} \sqrt{b^2-c}=a^2+b^2 \\
& \Rightarrow\left(a^2-c\right)\left(b^2-c\right)=c^2 \\
& \Rightarrow a^2 b^2-c b^2-c a^2+c^2=c^2 \\
& \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}
\end{aligned}
\)