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If \(3 x+y+k=0\) is a tangent to the circle \(x^2+y^2=10\), then \(k=\ldots . . .\).
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Verified Answer
The correct answer is:
\pm 5

Length of perpendicular
from \((0,0)\) on \(3 x+y+k=0\)
\(=\) radius of circle.
\(\Rightarrow\left|\frac{3 \times 0+0+k}{\sqrt{3^2+12}}\right|=\sqrt{10} \Rightarrow \frac{k}{\sqrt{10}}= \pm \sqrt{10}\)
\(\Rightarrow \quad k= \pm(\sqrt{10})^2 \Rightarrow k= \pm 10\)
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