Answer
$(x-3)^{2}+(y+1)^{2}=32$
Work Step by Step
Endpoints of a diameter are $P(-1,3)$ and $Q(7,-5)$
The equation of a circle is $(x-h)^{2}+(y-k)^{2}=r^{2}$, where $(h,k)$ is the center of the circle and $r$ is its radius.
Use the formula for the distance between two points to find the distance between the points given, which represents the diameter of the circle. The formula is $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
$d=\sqrt{(7+1)^{2}+(-5-3)^{2}}=\sqrt{(8)^{2}+(-8)^{2}}=\sqrt{64+64}=...$
$...=\sqrt{128}=8\sqrt{2}$
Since the radius of a circle is half its diameter, then the radius of the circle is:
$r=\dfrac{8\sqrt{2}}{2}=4\sqrt{2}$
Find the midpoint of the diameter $PQ$ to obtain the center of the circle:
$\Big(\dfrac{-1+7}{2},\dfrac{3-5}{2}\Big)=\Big(\dfrac{6}{2},\dfrac{-2}{2}\Big)=(3,-1)$
Since now both the center and the radius of the circle are known, substitute them into the equation of a circle formula:
$(x-h)^{2}+(y-k)^{2}=r^{2}$
$(x-3)^{2}+[y-(-1)]^{2}=(4\sqrt{2})^{2}$
$(x-3)^{2}+(y+1)^{2}=16(2)$
$(x-3)^{2}+(y+1)^{2}=32$