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?

Let $(a, b)$ be other end of the diameter then the center $(3, -2)$ is the mid-point of line joining the ends points of diameter $(a, b)$ & $(7, 2)$ 
Hence the coordinates of the center $(3, -2)$ are given as $$\left(\frac{a+7}{2}, \frac{b+2}{2}\right)\equiv(3, -2)$$ By comparing the corresponding coordinates, we get $$\frac{a+7}{2}=3\iff a=6-7=-1$$
$$\frac{b+2}{2}=-2\iff a=-4-2=-6$$ Hence the other end of the diameter is $(a, b)\equiv\color{blue}{ (-1, -6)}$
