Prove that the tangent at (3,-2) of the circle $x^2+y^2=13$touches the circle

$S=x^2+y^2+2x-10y-26=0$and find its point  of contact.

Tangent at (3,-2) to x² + y² = 13 
is S₁ = 0 → x (3) + y (-2) = 13
→ 3 x -2y -13 = 0 →(1)
S = x²+y² +2x-10y -26 = 0  →(2) 
center = (-1,5), radius = $\sqrt{1+25+26}$ 

                   =                  $\sqrt{52}$
                    =                 2 $\sqrt{13}$
d= perpendicular distance from (-1-5) to 3x- 2y -13 = 0
= $\frac{\left|-3-10-13\right|}{\sqrt[ ]{9+4}}=\frac{26}{\sqrt{13}}= 2\sqrt{13 }=r$

$\therefore$(1) is tangent to (2)
point  of contact = (h, k)= Foot of the perpendicular from (-1,5) to 13x-2y -13 =0
$\frac{\mathrm{h}+1}{3} = \frac{\mathrm{k}-5}{-2}= - \frac{(-3-10-13)}{9+4} = 2$
h = 5 k = 1
(h,k) = (5,1)