Coordinates and circles and triangles, oh my

Clearly the distance from $A_C$ to $B_C$ is $3$ , the distance from $A_C$ to $X$ is $\sqrt{5}$ , and the distance from $B_C$ to $X$ is $2$ . Thus, by Heron's Formula, the area of $\Delta A_C B_C X$ is $\sqrt{5}$ , and similarly, the area of $\Delta A_C B_C Y$ is $\sqrt{5}$ . Thus, since the quadrilateral $X A_C Y B_C$ is composed of those two triangles, the area is $2 \sqrt{5}$ and $2+5=\boxed{7}$ .

first find the centres between the circles it turns out to be 3. now draw the diagram u see that axb forms a right angled triangle at x (applying converse of pythagoras theorem ) same is the case for ayb .