Lengthy, Lengthier, Lengthiest!

Let a point of tangency to the circle be $(x, y)$ . The tangent to the circle must be perpendicular to the line that contains the same point and the radius of the circle. The slope of a tangent will be $\frac{y - 26}{x - 13}$ and the slope of the radius-containing line is $\frac{y - 6}{x - 3}$ . Because these lines are perpendicular, we have that $\frac{y - 26}{x - 13} = -\frac{x - 3}{y - 6} \longrightarrow x^2 + y^2 - 16x - 32y + 195 = 0 \hspace{.2cm} \text{ or } \hspace{.2cm} x^2 + y^2 = 16x + 32y - 195.$ Substituting this into the equation of the circle $\Omega$ given yields $10x + 20y - 600 = 0 \hspace{.2cm} \text{ or } \hspace{.2cm} x = 60 - 2y.$ Further substituting this into $\Omega$ gives us $5y^2 - 240y + 2880 = 0.$ The sum of the solutions to this $y_A + y_B$ is $\frac{240}{5} = 48$ . Similarly, we find $x_A + x_B = 24$ . Therefore $x_A + y_A + x_B + y_B = \boxed{72}$ .

Since the calculations are quite tedious , I will post only the method and not the working:

• First from the equation of circle we find that center $O \equiv (3,6)$ and radius $r=15\sqrt{2}$ .

• Let the slope of tangent be $\delta$ . So we have the equation of tangent :

$y-26=\delta(x-13) \Rightarrow \delta x-y -13(\delta-2)=0$

• Using distance of a line from a point formula we get:

$15\sqrt{2}= \dfrac{|\delta(3)-(6) -13(\delta-2)|}{\sqrt{\delta^2+1}} \\ \text{After squaring both sides and some simplification we have,}\\ \Rightarrow (7\delta+1)(\delta+1)=0 \\ \Rightarrow \delta = -\dfrac{1}{7} \ or \ -1$

• When $\delta = -\dfrac{1}{7}$ , we have equation of $t_1$ as : $x+7y=195$ . To get the coordinates of $A$ we solve the following system of equation:

$\begin{array}{c}& x+7y & = & 195 \\ x^2+y^2-6x-12y-405 & = & 0 \end{array}$

After solving , we get $(x_A,y_A )= (6,27)$ .

• When $\delta = -1$ , we have equation of $t_2$ as : $x+y=39$ . To get the coordinates of $B$ we solve the following system of equation:

$\begin{array}{c}& x+y & = & 39 \\ x^2+y^2-6x-12y-405 & = & 0 \end{array}$

After solving , we get $(x_B,y_B )= (18,21)$ .

So required sum $= 6+27+18+21= \boxed{72}$ .