Question

Find the equation of a circle in standard form where C(6, -2) and D(-4, 4) are endpoints of a diameter.

Answer 1

center is the midpoint. you find the midpoint via
\[(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\] in other words take the average in each coordinate. you might could do it in your head, but we can write it out

Answer 2

\[(\frac{6-4}{2},\frac{-2+4}{2})\]
\[(\frac{2}{2},\frac{2}{2})\]
\[(1,1)\] so that is the center

Answer 3

and your equation will look like
\[(x-1)^2+(y-1)^2=r^2\] so all we need is
\[r^2\]

Answer 4

we get that by finding the distance between
\[(1,1)\] and either one of the other points. i choose
\[(6,-2)\] and so
\[r^2=(6-1)^2+(1+2)^2=5^2+3^2=25+9=36\] and your equation is
\[(x-1)^2+(y-1)^2=36\]