Let the equation of line be $y = mx + c$
Since this line is tangent to the circle
$x^{2} +y^{2} = 5$
$ \therefore c =\pm a \sqrt{1+m^{2}} $
$=\pm \sqrt{5} \sqrt{1+m^{2}} \,\,\,\,\,\, \dots(i)$
Also, the above line is tangent to the parabola
$ y^{2} =40\,x$
$ \therefore c = \frac{a}{m} = \frac{10}{m}\,\,\,\,\,\,\dots(ii)$
From Eqs. (i) and (ii)
$\Rightarrow \frac{10}{m} = \pm\,\sqrt{5}\,\sqrt{1+m^{2}} $
$\Rightarrow \frac{20}{m^{2}} = 1+m^{2} $
$ \Rightarrow m^{4} +m^{2} - 20 = 0$
$\Rightarrow \left(m^{2} + 5\right) \left(m^{2}-4\right) = 0$
$ \Rightarrow m^{2} =4 , m^{2} \ne - 5$
$ \Rightarrow m = \pm 2 $
$ \Rightarrow c = \pm 5 $
$ \therefore y = \pm 2 x \pm 5 $