Answer
$(x-2)^2+(y-5)^2=25$
Work Step by Step
The center of the circle is the midpoint of the diameter.
Find the center (midpoint of the diameter) using the midpoint formula $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$ to obtain:
center = $\left(\dfrac{-1+5}{2}, \dfrac{1+9}{2}\right)=\left(\dfrac{4}{2}, \dfrac{10}{2}\right)=(2, 5)$
RECALL:
The standard form of a circle whose center is at $(h, k)$ and radius $r$ is $(x--h)^2+(y-k)^2=r^2$.
The center is at $(2, 5)$, so $h=2$ and $k=5$.
Thus, the standard form of the given circle's equation of the circle is:
$(x-2)^2+(y-5)^2=r^2$
The circle passes through the point $(5, 9)$.
This means that the coordinates of this point satisfy the equation of the circle.
Substitute the x and y values of the point into the equation to obtain:
$(x-2)^2+(y-5)^2=r^2
\\(5-2)^2+(9-5)^2=r^2
\\3^2+4^2=r^2
\\9+16=r^2
\\25=r^2$
Thus, the standard form of the given circle's equation is:
$(x-2)^2+(y-5)^2=25$
