If the circles $x^2+y^2+2ax+c^2=0$ and $x^2+y^2+2by+c^2=0$ touch externally, prove that $\frac {1}{a^2} +\frac {1}{b^2}=\frac {1}{c^2}$.
My Attempt
Here
$$x^2+y^2+2ax+c^2=0$$ $$x^2+2.x.a+a^2-a^2+y^2+c^2=0$$ $$(x+a)^2=a^2-c^2-y^2$$
Then what should I do next?
Centres: $A(-a,0)$ and $B(0,-b)$
Radii: $r_{1}=\sqrt{a^2-c^2}$, $r_{2}=\sqrt{b^2-c^2}$
Let $O=(0,0)$, then $OA \perp OB$.
By Pythagoras' Theorem:
\begin{align*} OA^2+OB^2 &= AB^2 \\ a^2+b^2 &= (r_{1}+r_{2})^{2} \\ a^2+b^2 &= a^2+b^2-2c^2+2\sqrt{(a^2-c^2)(b^2-c^2)} \\ c^4 &= (a^2-c^2)(b^2-c^2) \\ (a^2+b^2)c^2 &= a^2 b^2 \\ \frac{1}{a^2}+\frac{1}{b^2} &= \frac{1}{c^2} \end{align*}
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} x^{2} + y^{2} + 2ax + c^{2} & = 0\tag{1} \\[3mm] x^{2} + y^{2} + 2by + c^{2} & = 0\tag{2} \end{align} By substracting both members of $\pars{1}\ \mbox{and}\ \pars{2}$ we conclude $\ds{y = {a \over b}\,x}$. Replacing, for example, in $\pars{1}$: $$ \pars{b^{2} + a^{2}}x^{2} + 2abx + c^{2} = 0 $$ This is a second degree equation for $x$. Since there is just one intersection: $$ 0 = \pars{2ab}^{2} - 4\pars{b^{2} + a^{2}}\pars{c^{2}} $$ Dividing by $\ds{4a^{2}b^{2}c^{2}}$: $$ \color{#f00}{0 = {1 \over c^{2}} - {1 \over a^{2}} - {1 \over b^{2}}} $$
