Locate The A B

Since $AC = AO$ , we have:

$\begin{aligned} AC^2 & = AO^2 \\ \left(x_a - \frac {24}5 \right)^2 + \left(y_a + \frac {12}5 \right)^2 & = \left(x_a - 0 \right)^2 + \left(y_a - 0 \right)^2 \\ x_a^2 - \frac {48}5 x_a + \frac {576}{25} + y_a^2 + \frac {24}5 y_a + \frac {144}{25} & = x_a^2 + y_a^2 \\ - \frac {48}5 x_a + \frac {24}5 y_a + \frac {720}{25} & = 0 \\ - {\color{#3D99F6}4 x_a} + 2 y_a + 12 & = 0 & \small \color{#3D99F6} \text{Note that } 4x_a + 3y_a - 12 = 0 \\ {\color{#3D99F6}3y_a - 12} + 2 y_a + 12 & = 0 \\ \implies y_a & = 0 \\ x_a & = 3 \end{aligned}$

Therefore, $A(0,3)$ is on the $x$ -axis and $OA$ is along the $x$ -axis. Excircle centered at $K$ is hence tangent to the $x$ -axis and has a radius of 6 and since the excenter $K$ is at $(6,6)$ the excircle is also tangent to the $y$ -axis as shown in the figure below. Implying that the line $OB$ is along $y$ -axis and $x_b = 0$ , $\implies y_b = 4$ .

Therefore, $x_a+y_a+x_b+y_b = 3+0+0+4 = \boxed{7}$ .

The distance from $K(6,6)$ to the line $4x + 3y - 12 = 0$ is given by $\frac{|4 \cdot 6 + 3 \cdot 6 - 12|}{\sqrt{4^2 + 3^2}} = 6,$ which is also the radius of the circle.

Since the center of the circle is $K(6,6)$ , and its radius is 6, it is tangent to both the $x$ -axis and $y$ -axis. Thus, $A$ and $B$ are the $x$ -intercept and the $y$ -intercept of the line $4x + 3y - 12 = 0$ , in some order. The $x$ -intercept is (3,0) and the $y$ -intercept is (0,4), so the answer is $3 + 0 + 0 + 4 = 7$ .

Note that information about point $C$ is never used, so it is not needed.

A lies on perpendicular bisector of OC as well as on line d. Solve the two equations to get A(3,0).

Now, reflect A about line OK to get point X. Note that this point X lies on line segment OB. Hence find equation of OB. Solve it with equation of line d to get B(0,4)

(See above figure for details)

There are a few approaches to this problem but this is one i have come up with.

First we need to find some geometric proofs.

Let point $D \in (d)$ so that $BO=BD$ and point $B$ lies between points $A$ and $D$ . Let $E=AK \cap OC, F=BK \cap OD$ and $KH \perp (d)$ with $H \in (d)$ .

Since point $K$ is the excenter of $O$ so $KE$ is the bisector of $\angle OAC$ . We also have $\Delta OAC$ is an isosceles triangle so we can imply that point $E$ is the midpoint of $OC$ and $KC=KO$ .

Similarly, point $F$ is the midpoint of $OD$ and $KD=KO$ .

We have $KC=KO=KD$ so $\Delta CKD$ is an isosceles triangle. Therefore, point $H$ is the midpoint of $CD$ .

Then, let's take these proofs into the coordinates system.

We have had the coordinates of point $O$ and $C$ so we can find the coordinates of midpoint $E$ of $OC$ that is $(\frac{12}{5};\frac{-6}{5})$ . Line segment $EK \perp OC$ and passes point $E$ so its equation is $2x-y-6=0$ .

We have $A= EK \cap (d)$ and both line equations have been found so we can find the coordinates of point $A$ that is $(3;0)$

We have line segment $KH \perp (d)$ and passes point $K$ so its equation is $3x-4y+6=0$ . We also have point $H=KH \cap (d)$ so its coordinates is $(\frac{6}{5};\frac{12}{5})$ . Since we have proved that $H$ is the midpoint of $CD$ so we can find the coordinates of point $D$ that is $(\frac{-12}{5};\frac{36}{5})$

By the same way that we have found the coordinates of point $A$ , we can find the coordinates of point $B$ that is $(0;4)$ .

In conclusion, $x_a+y_a+x_b+y_b = \boxed7$

$The ~line~ d ~is~\dfrac X 3+\dfrac Y 4 = 1.~So~X~and~Y~intersects~are~3~and~4.\\ K~ is~on~ Y=X, ~that~~bisects~the~right~\angle~ through~O,~\therefore~A~and~B~are~each~on~one~of~the~axis.\\ C~is~~near~A,~and~in~fourth~quadrant.\\ \therefore~A, ~is~at~intersection~of~Y=0~and~d,~\therefore~A(3,0)=(X_a,Y_a).~~~~~Similarly~ B(0,4)=(X_b,Y_b).\\ \therefore~X_a+Y_a+X_b+Y_b=3+0+0+4=\Large~~\color{#D61F06}{7}.\\$