If \(3 x+y+k=0\) is a tangent to the circle \(x^{2}+y^{2}=10\), the values of \(k\) are

Given, line is \(3 x+y+k=0\)
\[
\Rightarrow \quad y=-3 x-k
\]
And equation of circle is \(x^{2}+y^{2}=10\)
Here, \(a^{2}=10, m=-3, c=-k\)
If given line touches the circle, then length of intercept \(=0\)
\(\Rightarrow \quad 2 \sqrt{\frac{a^{2}\left(1+m^{2}\right)-c^{2}}{1+m^{2}}}=0\)
\(\Rightarrow \quad 2 \sqrt{\frac{10(1+9)-k^{2}}{1+9}}=0\)
\(\Rightarrow \quad \sqrt{100-k^{2}}=0\)
\(\Rightarrow \quad 100-k^{2}=0 \Rightarrow k=\pm 10\)
Alternative : If the given line is tangent to the circle, then the length of the perpendicular from the centre upon the line is equal to the radius of the circle.
\[
\begin{aligned}
&\text { i.e. } \quad\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|=r \\
&\Rightarrow \quad\left|\frac{3 \times 0+6 \times 0+k}{\sqrt{(3)^{2}+(1)^{2}}}\right|=\sqrt{10} \\
&\Rightarrow \quad\left|\frac{k}{\sqrt{10}}\right|=\sqrt{10} \\
&\Rightarrow \quad k=\sqrt{100} \Rightarrow k=\pm 10
\end{aligned}
\]