Answer
$(x-2)^{2}+(y-5)^{2}=25$
Work Step by Step
Endpoints of a diameter are $P(-1,1)$ and $Q(5,9)$
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 $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$
$d=\sqrt{(5+1)^{2}+(9-1)^{2}}=\sqrt{6^{2}+8^{2}}=\sqrt{36+64}=...$
$...=\sqrt{100}=10$
Since the radius of a circle is half its diameter, then the radius of the circle is:
$r=\dfrac{10}{2}=5$
Find the midpoint of the diameter $PQ$ to obtain the center of the circle:
$\Big(\dfrac{-1+5}{2},\dfrac{1+9}{2}\Big)=\Big(\dfrac{4}{2},\dfrac{10}{2}\Big)=(2,5)$
Since now both the center and the radius of the circle are known, substitute them into the equation of a circle formula:
$(x-h)^{2}+(y-k)^{2}=r^{2}$
$(x-2)^{2}+(y-5)^{2}=5^{2}$
$(x-2)^{2}+(y-5)^{2}=25$
