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 $(a, b)\equiv\color{blue}{ (-1, -6)}$
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 $b$).
