Explanation

(a)
The equation of circles whose radius is $r$ and centres (2, 3) and (5, 6) is
$(x - 2)^{2} + (y - 3)^{2} = r^{2}$
And $(x - 5)^{2} + (y - 6)^{2} = r^{2}$
$\Rightarrow \ \ \ x^{2} + y^{2} - 4x - 6y + \left( - r^{2} + 13 \right) = 0$
And $x^{2} + y^{2} - 10x - 12y + \left( - r^{2} + 61 \right) = 0$
Since, circles cut orthogonally, then
$2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}$
$\Rightarrow \ 2(2)(5) + 2(3)(6) = 13 - r^{2} + 61 - r^{2}$
$\Rightarrow \ 2r^{2} = 18\ \ \ \Rightarrow \ \ \ r = 3$