Answer
$\color{blue}{\bf{(3,1)}}$
Work Step by Step
If using a graphing calculator, we'll start by writing each seismic stations center and radius in $\bf\text{center radius form:}$
$$\bf{(x-h)^2+(y-k)^2=c}$$
$\bf{\text{where }(h,k)\text{ is the center, and c is the radius squared.}}$
$\bf{\text{ we'll solve for y and graph each circle.}}$
$\bf{\text{Station X has its center at }(7,4)\text{and a radius of }5}$
$(x-7)^2+(y-4)^2=5^2$
$(x-7)^2+(y-4)^2=25$
$(y-4)^2=25-(x-7)^2$
$y-4=\pm\sqrt{25-(x-7)^2}$
$\bf{y=4+\sqrt{25-(x-7)^2}}$ and $\bf{y=4-\sqrt{25-(x-7)^2}}$
which we can graph in our calculator
$\bf{\text{Station Y has its center at }(-9,-4)\text{and a radius of }13}$
$(x-(-9))^2+(y-(-4))^2=13^2$
$(x+9)^2+(y+4)^2=169$
$(y+4)^2=169-(x+9)^2$
$y+4=\pm\sqrt{169-(x+9)^2}$
$y=-4\pm\sqrt{169-(x+9)^2}$
$\bf{y=-4+\sqrt{169-(x+9)^2}}$ and $\bf{y=-4-\sqrt{169-(x+9)^2}}$
which we can graph in our calculator
$\bf{\text{Station Z has its center at }(-3,9)\text{and a radius of }10}$
$(x-(-3))^2+(y-9)^2=10^2$
$(x+3)^2+(y-9)^2=100$
$(y-9)^2=100-(x+3)^2$
$y-9=\pm\sqrt{100-(x+3)^2}$
$y=9\pm\sqrt{100-(x+3)^2}$
$\bf{y=9+\sqrt{100-(x+3)^2}}$ and $\bf{y=9-\sqrt{100-(x+3)^2}}$
which we can graph in our calculator
After graphing the three circles as below we see that the epicenter is at $\color{blue}{\bf{(3,1)}}$
