If the lengths of the tangents drawn from $P$ to the circles $x^2+y^2-2 x+4 y-20=0$ and $x^2+y^2-2 x-8…
If the lengths of the tangents drawn from $P$ to the circles $x^2+y^2-2 x+4 y-20=0$ and $x^2+y^2-2 x-8 y+1=0$ are in the ratio $2: 1$, then the locus $P$ is
$x^2+y^2+2 x+12 y+8=0$
$x^2+y^2-2 x+12 y+8=0$
$x^2+y^2+2 x-12 y+8=0$
$x^2+y^2-2 x-12 y+8=0$
Solution
We know that, length of tangent drawn from $\left(x_1, y_1\right)$ to the circle $x^2+y^2+2 g x+2 f y+c=0$ is $\sqrt{x_1^2+y_1^2+2 g x_1+2 f y_1+c}$
Let $P(h, k)$
$\therefore$ According to the question,
$
\begin{aligned}
& \frac{\sqrt{h^2+k^2-2 h+4 k-20}}{\sqrt{h^2+k^2-2 h-8 k+1}}=\frac{2}{1} \\
\Rightarrow & h^2+k^2-2 h+4 k-20=4\left(h^2+k^2-2 h-8 k+1\right) \\
\Rightarrow & h^2+k^2-2 h+4 k-20=4 h^2+4 k^2-8 h-32 k+4 \\
\Rightarrow & 3 h^2+3 k^2-6 h-36 k+24=0 \\
\Rightarrow & h^2+k^2-2 h-12 k+8=0
\end{aligned}
$
$\therefore$ Locus of point $P$ is $x^2+y^2-2 x-12 y+8=0$