Explanation

(d)
Given equation of circles are $x^{2} + y^{2} - 2x - 4y + 1 = 0$ and $x^{2} + y^{2} - 12x - 16y + 91 = 0$ whose centre and radius are $C_{1}(1,\ 2),\ r_{1} = 2$ and $C_{2}(6,\ 8),\ r_{2} = 3$
$\therefore C_{1}C_{2} = \sqrt{(1 - 6)^{2} + (2 - 8)^{2}}$
$= \sqrt{25 + 36} = \sqrt{61}$
And $r_{1} + r_{2} = 2 + 3 = 5$
$\because C_{1}C_{2} > r_{1} + r_{2}$
$\therefore$ Number of common tangents =4