## Precalculus (6th Edition) Blitzer

a. $(5,10)$ b. $\sqrt 5$ c. $(x-5)^2+(y-10)^2=5$
a. Given the coordinates of the ends of a diameter: $(3,9)$ and $(7,11)$, we can find the center of the circle as $(\frac{7+3}{2},\frac{11+9}{2})$ or $(5,10)$ b. The radius is half of the diameter; we have $r=\frac{\sqrt {(7-3)^2+(11-9)^2}}{2}=\frac{\sqrt {16+4}}{2}=\sqrt 5$ c. Based on the above results, we can write the standard form of the circle’s equation as $(x-5)^2+(y-10)^2=5$