Answer
$\color{blue}{\bf{(5,2)}}$
Work Step by Step
If using a graphing calculator, we'll start by writing each seismic stations center and radius in $\bf\text{center radius form:}$
$$\bf{(x-h)^2+(y-k)^2=c}$$
$\bf{\text{where }(h,k)\text{ is the center, and c is the radius squared.}}$
$\bf{\text{ we'll then solve for y and graph each circle.}}$
$\bf{\text{Station P has its center at }(3,1)\text{ and a radius of }\sqrt{5}}$
$(x-3)^2+(y-1)^2=\sqrt{5}^2$
$(y-1)^2=5-(x-3)^2$
$y-1=\pm\sqrt{5-(x-3)^2}$
$y=1\pm\sqrt{5-(x-3)^2}$
$\bf{y=1+\sqrt{5-(x-3)^2}}$ and $\bf{y=1-\sqrt{5-(x-3)^2}}$
which we can graph in our calculator
$\bf{\text{Station Q has its center at }(5,-4)\text{ and a radius of }6}$
$(x-5)^2+(y-(-4))^2=6^2$
$(x-5)^2+(y+4)^2=36$
$(y+4)^2=36-(x-5)^2$
$y+4=\pm\sqrt{36-(x-5)^2}$
$y=-4\pm\sqrt{36-(x-5)^2}$
$\bf{y=-4+\sqrt{36-(x-5)^2}}$ and $\bf{y=-4-\sqrt{36-(x-5)^2}}$
which we can graph in our calculator
$\bf{\text{Station R has its center at }(-1,4)\text{ and a radius of }2\sqrt{10}}$
$(x-(-1))^2+(y-4)^2=(2\sqrt{10})^2$
$(x+1)^2+(y-4)^2=4(10)$
$(y-4)^2=4(10)-(x+1)^2$
$y-4=\pm\sqrt{4(10)-(x+1)^2}$
$y=4\pm\sqrt{4(10)-(x+1)^2}$
$\bf{y=4+\sqrt{4(10)-(x+1)^2}}$ and $\bf{y=4-\sqrt{4(10)-(x+1)^2}}$
which we can graph in our calculator
After graphing the three circles as below we see that the epicenter is at $\color{blue}{\bf{(5,2)}}$
