Question

A diameter of a circle has endpoints P(-10,-2) and Q(4,6).

Answer 1

Check the picture below.

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ P(\stackrel{x_1}{-10}~,~\stackrel{y_1}{-2})\qquad Q(\stackrel{x_2}{4}~,~\stackrel{y_2}{6}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{4-10}{2}~~,~~\cfrac{6-2}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{4}{2} \right)\implies \stackrel{\textit{center}}{(-3~,~2)} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{\textit{center}}{(\stackrel{x_1}{-3}~,~\stackrel{y_1}{2})}\qquad Q(\stackrel{x_2}{4}~,~\stackrel{y_2}{6})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{[4-(-3)]^2+[6-2]^2}\implies r=\sqrt{(4+3)^2+(6-2)^2} \\\\\\ r=\sqrt{49+16}\implies r=\sqrt{65} \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{2}{ k})\qquad \qquad radius=\stackrel{\sqrt{65}}{ r} \\[2em] [x-(-3)]^2+[y-2]^2=(\sqrt{65})^2\implies (x+3)^2+(y-2)^2=65$