The equation of common tangent to the circle \(x^{2}+y^{2}=2\) and the parabola \(y^{2}=8 x\) is \(x+y-k\). Then value of \(k\) is

Given parabola is, \(y^{2}=8 x\)
Here,
\(
4 a=8 \Rightarrow a=2
\)
Any tangent to the parabola is \(y=m x+\frac{a}{m}\)
or
\(
m x-y+\frac{2}{m}=0
\)
If it is a tangent to the circle \(x^{2}+y^{2}=2\), then length of perpendicular from centre \((0,0)\) is equal to radius \(\sqrt{2}\).
\(\therefore \frac{2 / m}{\sqrt{m^{2}+1}}=\sqrt{2} \)
\( \Rightarrow \frac{4}{m^{2}}=2\left(m^{2}+1\right) \)
\( \Rightarrow m^{4}+m^{2}-2=0 \)
\( \Rightarrow \left(m^{2}+2\right)\left(m^{2}-1\right)=0\)
\(\Rightarrow m^{2}=\pm 1\)
Hence, the common tangent are \(y=\pm(x+2)\). But given equation of tangent is \(x+y=k\) On comparing, we get \(k=-2\)