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If $3 x+y+k=0$ is a tangent to the circle $x^{2}+y^{2}=10$, the values of $k$ are
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Verified Answer
The correct answer is:
$\pm 10$
We have, $3 x+y+k=0$ is iargent to the circle $x^{2}+v^{2}=10 .$
So, distance of line $3 x+y+k=0$ from centre of the circle will he equal to the radius of circle.
$$
\begin{aligned}
&\therefore \quad \mid \frac{3(0)+(0)+K}{\sqrt{(3)^{2}+(1)^{2}} \mid}=\sqrt{10} \\
&\Rightarrow \quad \frac{K}{\sqrt{10}}=\pm \sqrt{10} \Rightarrow K=\pm 10
\end{aligned}
$$
So, distance of line $3 x+y+k=0$ from centre of the circle will he equal to the radius of circle.
$$
\begin{aligned}
&\therefore \quad \mid \frac{3(0)+(0)+K}{\sqrt{(3)^{2}+(1)^{2}} \mid}=\sqrt{10} \\
&\Rightarrow \quad \frac{K}{\sqrt{10}}=\pm \sqrt{10} \Rightarrow K=\pm 10
\end{aligned}
$$
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