Show that locus of poles of the tangent to the circle x² + y² = a² w.r.t the circle $(\mathrm{x}+\mathrm{a}{)}^2+{\mathrm{y}}^2=2{\mathrm{a}}^2\text{ is }{\mathrm{y}}^2+4\mathrm{ax}=0$

Given circles are ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2 ...\left(1\right)$
$\text{ and }{\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{ax}-{\mathrm{a}}^2=0 ...\left(2\right)$
Let $\mathrm{\ell }$ be a tangent to circle (1)

$\text{ Let }\mathrm{P}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\text{ be the pole of }\mathrm{\ell }\text{ w.r.t }{2}^{\text{nd }}\text{ circle }$
$\text{ Equation of }\mathrm{\ell }\text{ is }{\mathrm{S}}_1=0\left(\text{ wrt }{2}^{\text{nd }}\text{ circle }\right)$
$\begin{array}{l}\Rightarrow {\mathrm{xx}}_1+{\mathrm{yy}}_1+\mathrm{a}\left(\mathrm{x}+{\mathrm{x}}_1\right)-{\mathrm{a}}^2=0\\ \Rightarrow {\mathrm{xx}}_1+{\mathrm{yy}}_1+\mathrm{ax}+{\mathrm{ax}}_1-{\mathrm{a}}^2=0\\ \Rightarrow \left({\mathrm{x}}_1+\mathrm{a}\right)\mathrm{x}+{\mathrm{yy}}_1+\mathrm{a}\left({\mathrm{x}}_1-\mathrm{a}\right)=0\end{array}$
But, it is a tangent to 1st circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$
$\begin{array}{l}\Rightarrow \mathrm{d}=\mathrm{r}\Rightarrow \frac{\left|\mathrm{a}\left({\mathrm{x}}_1-\mathrm{a}\right)\right|}{\sqrt{{\left({\mathrm{x}}_1+\mathrm{a}\right)}^2+{\mathrm{y}}_1^2}}=\mathrm{a}\\ \Rightarrow \left|{\mathrm{x}}_1-\mathrm{a}\right|=\sqrt{{\left({\mathrm{x}}_1+\mathrm{a}\right)}^2+{\mathrm{y}}_1^2}\end{array}$
$\text{ S.O.B.S }{\left({\mathrm{x}}_1-\mathrm{a}\right)}^2={\left({\mathrm{x}}_1+\mathrm{a}\right)}^2+{\mathrm{y}}_1^2$
$\Rightarrow {\mathrm{y}}_1^2+{\left({\mathrm{x}}_1+\mathrm{a}\right)}^2-{\left({\mathrm{x}}_1-\mathrm{a}\right)}^2=0\Rightarrow {\mathrm{y}}_1^2+4\mathrm{ax}=0$
$\text{ Locus of }\mathrm{P}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\text{ is }{\mathrm{y}}^2+4\mathrm{ax}=0$