Finding the other end of the Diameter For a National Board Exam Review

A circle has it center at $(3,-2)$ and one end of a diameter at $(7,2)$. Find the other end of the diameter.

Answer is $(-1,6)$
$$m=\frac{y^2-y^1}{x^2-x^1}=\frac{2-(-2) }{7-3}$$
$$=(3-7)^2+(-2-2)^2=r^2=32$$
$$r =\sqrt{32}$$
Plugin $(7,2)$ into
$$y = mx+b$$
$$b = -5$$
Solve two linear equations:
$$32 = (x-h)^2 + (y-k)^2$$
$$32 = (x-7)^2 + ((x-5)-2)^2$$
$$y=x-5$$
I get $(3,-2)$. What am I doing wrong? Any hint?

Here's another approach:
The center of a circle is the midpoint of any diameter.  If the coordinates of the unknown point are $(a,b)$, using the midpoint formula:
$\frac{a+7}{2}=3$, and...(see if you can work out the equation for $b$).
