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. $$ \begin{aligned} &\text { i.e. } \quad\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\right|=r \\ &\Rightarrow \quad\left|\frac{3 \times 0+6 \times 0+k}{\sqrt{(3)^{2}+(1)^{2}}}\right|=\sqrt{10} \\ &\Rightarrow \quad\left|\frac{k}{\sqrt{10}}\right|=\sqrt{10} \\ &\Rightarrow \quad k=\sqrt{100} \Rightarrow k=\pm 10 \end{aligned} $$