Algebra and Trigonometry 10th Edition

$(x-7)^2+(y-5)^2=116$
The middlepoint between $(11,-5)$ and $(3,15)$ is the center of the circle: $\frac{(11,-5)+(3,15)}{2}=(7,5)$, that is: $h=7$, $k=5$ The distance from the center to an endpoint is the radius: $r=\sqrt {(7-11)^2+[5-(-5)]^2}=\sqrt {16+100}=\sqrt {116}$ Equation of a circle: $(x−h)^2+(y−k)^2=r^2$ (standard form) $(x-7)^2+(y-5)^2=(\sqrt {116})^2$ $(x-7)^2+(y-5)^2=116$