Answer
$(x-3)^2+(y+1)^2=32$
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 = $\left(\dfrac{-1+7}{2}, \dfrac{3+(-5)}{2}\right)=\left(\dfrac{6}{2}, \dfrac{-2}{2}\right)=(3, -1)$
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 $(3, -1)$, so $h=3$ and $k=-1$.
Thus, the standard form of the given circle's equation of the circle is:
$(x-3)^2+[y-(-1)]^2=r^2
\\(x-3)^2+(y+1)^2=r^2$
The circle passes through the point $(-1, 3)$.
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:
$(x-3)^2+(y+1)^2=r^2
\\(-1-3)^2+(3+1)^2=r^2
\\(-4)^2+4^2=r^2
\\16+16=r^2
\\32=r^2$
Thus, the standard form of the given circle's equation is:
$(x-3)^2+(y+1)^2=32$
