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If the length of the tangent from $(h, k)$ to the circle $x^2+y^2=16$ is twice the length of the tangent from the same point to the circle $x^2+y^2+2 x+2 y=0$, then
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Verified Answer
The correct answer is:
$3 h^2+3 k^2+8 h+8 k+16=0$
Given equations of circle is
$$
x^2+y^2=16
$$
and $\quad x^2+y^2+2 x+2 y=0$
According to the questions,
Length of the tangent from $(h, k)$ to the circle
$$
x^2+y^2=16
$$
$=2 \times$ Length of the tangent from $(h, k)$ to the circle $x^2+y^2+2 x+2 y=0$
$$
\Rightarrow \sqrt{h^2+k^2-16}=2 \times \sqrt{h^2+k^2+2 h+2 k}
$$
On squaring both sides, we get
$$
\begin{aligned}
& h^2+k^2-16=4 \times\left(h^2+k^2+2 h+2 k\right) \\
\Rightarrow \quad & 3 h^2+3 k^2+8 h+8 k+16=0
\end{aligned}
$$
$$
x^2+y^2=16
$$
and $\quad x^2+y^2+2 x+2 y=0$
According to the questions,
Length of the tangent from $(h, k)$ to the circle
$$
x^2+y^2=16
$$
$=2 \times$ Length of the tangent from $(h, k)$ to the circle $x^2+y^2+2 x+2 y=0$
$$
\Rightarrow \sqrt{h^2+k^2-16}=2 \times \sqrt{h^2+k^2+2 h+2 k}
$$
On squaring both sides, we get
$$
\begin{aligned}
& h^2+k^2-16=4 \times\left(h^2+k^2+2 h+2 k\right) \\
\Rightarrow \quad & 3 h^2+3 k^2+8 h+8 k+16=0
\end{aligned}
$$
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