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If $P\left(x_1, y_1\right)$ is a point such that the length of the tangents from it to the circles $x^2+y^2-4 x-6 y-12=0$ and $x^2+y^2+6 x+18 y+26=0$ are in the ratio $2: 3$, then the locus of $P$ is
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Verified Answer
The correct answer is:
$5x^2+ 5y^2-60x-126y-212=0$
Given equations of circle are
$$
x^2+y^2-4 x-6 y-12=0
$$
and $x^2+y^2+6 x+18 y+26=0$
Tangents from $P\left(x_1, y_1\right)$ to the circles are in the ratio of $2: 3$.
$$
\begin{aligned}
& \text { So, } \frac{\sqrt{x_1^2+y_1^2-4 x_1-6 y_1-12}}{\sqrt{x_1^2+y_1^2+6 x_1+18 y_1+26}}=\frac{2}{3} \\
& \Rightarrow \quad \frac{x_1^2+y_1^2-4 x_1-6 y_1-12}{x_1^2+y_1^2+6 x_1+18 y_1+26}=\frac{4}{9} \\
& \Rightarrow 5 x_1^2+5 y_1^2-60 x_1-126 y_1-212=0
\end{aligned}
$$
Then, the locus of $P\left(x_1, y_1\right)$ is
$$
5 x^2+5 y^2-60 x-126 y-212=0
$$
No option is correct.
$$
x^2+y^2-4 x-6 y-12=0
$$
and $x^2+y^2+6 x+18 y+26=0$
Tangents from $P\left(x_1, y_1\right)$ to the circles are in the ratio of $2: 3$.
$$
\begin{aligned}
& \text { So, } \frac{\sqrt{x_1^2+y_1^2-4 x_1-6 y_1-12}}{\sqrt{x_1^2+y_1^2+6 x_1+18 y_1+26}}=\frac{2}{3} \\
& \Rightarrow \quad \frac{x_1^2+y_1^2-4 x_1-6 y_1-12}{x_1^2+y_1^2+6 x_1+18 y_1+26}=\frac{4}{9} \\
& \Rightarrow 5 x_1^2+5 y_1^2-60 x_1-126 y_1-212=0
\end{aligned}
$$
Then, the locus of $P\left(x_1, y_1\right)$ is
$$
5 x^2+5 y^2-60 x-126 y-212=0
$$
No option is correct.
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