Explanation

(c)
Centre is $(7,\ - 1)$ and radius is 5
Let $y = mx$ be the tangent on the circle
$\therefore$ length of perpendicular from centre is equal to the radius of circle
$\Rightarrow \ \ \ \frac{7m + 1}{\sqrt{1 + m^{2}}} = \pm 5$
$\Rightarrow \ \ \ 49m^{2} + 1 + 14m\text{=}25\left( 1 + m^{2} \right)$
$\Rightarrow \ \ \ 12m^{2} + 7m - 12 = 0$
$\Rightarrow \ \ \ (3m + 4)(4m - 3) = 0$
$\Rightarrow \ \ \ m_{1} = - \frac{4}{3}\ \ and\ \ \ m_{2} = \frac{3}{4}$
$\therefore\ \ \ m_{1}m_{2} = - \frac{4}{3}.\frac{3}{4} = - 1$
Hence, tangents are perpendicular to each other
Alternate $\theta = 2\tan^{- 1}\frac{r}{\sqrt{S_{1}}}$
$= 2\tan^{- 1}{\frac{5}{5} =}\frac{\pi}{2}$