The poles of the tangents to the circle x 2 + y 2 = 4 with respect to the circle ( x + 2 ) 2 + y 2 = 8 lie on

Given: $x^2+y^2=4$

${\left(x+2\right)}^2+y^2=8$

Equation of tangent at any point on the circle $x^2+y^2=4$is

$x\cos \theta +y\sin \theta =r$

$x\cos \theta +y\sin \theta =2 ...\left(1\right)$

The polar of $\left(x_1,y_1\right)$w.r.t. ${\left(x+2\right)}^2+y^2=8$

$x{x}_1+y{y}_1+2\left(x+x_1\right)-4=0$

$\Rightarrow \left(x_1+2\right)x+y_{1}y+2\left(x_1-2\right)=0 ...\left(2\right)$

from equation $\left(1\right)$& $\left(2\right)$, we get

$\frac{\cos \theta }{x_1+2}=\frac{\sin \theta }{y_1}=\frac{-2}{2\left(x_1-2\right)}$

$\cos \theta =-\frac{\left(x_1+2\right)}{\left(x_1-2\right)}$& $\sin \theta =-\frac{y}{x_1-2}$

${\cos }^2\theta +{\sin }^2\theta =\frac{{\left(x_1+2\right)}^2}{{\left(x_1-2\right)}^2}+\frac{{y_1}^2}{{\left(x_1-2\right)}^2}$

$\Rightarrow \frac{{\left(x_1+2\right)}^2+{y_1}^2}{{\left(x_1-2\right)}^2}=1$

$\Rightarrow {x_1}^2+4x_1+4+{y_1}^2={x_1}^2-4x_1+4$

$\Rightarrow {y_1}^2+8x_1=0$

$\Rightarrow y^2+8x=0$