Question

The diameter of a circle has endpoints (-10,5) and (-4,11)

Answer 1

well, since we know the endpoints for the diameter, its midpoint will be where the center of the circle is located, so

$~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{11}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -4 -10}{2}~~~ ,~~~ \cfrac{ 11 + 5}{2} \right)\implies \left(\cfrac{-14}{2}~~,~~\cfrac{16}{2} \right)\implies \stackrel{center}{(-7~~,~~8)}$

and if we get the distance between those endpoints, and take half of that, that'd be the radius of it.

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{11})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-4 - (-10)]^2 + [11 - 5]^2}\implies d=\sqrt{(-4+10)^2+6^2} \\\\\\ d=\sqrt{6^2+6^2}\implies d=\sqrt{72}~\hfill \stackrel{radius=half~that}{\cfrac{\sqrt{72}}{2}} \\\\[-0.35em] ~\dotfill$

$\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-7}{ h},\stackrel{8}{ k})\qquad \qquad radius=\stackrel{\frac{\sqrt{72}}{2}}{ r} \\\\\\\ [x-(-7)]^2~~ + ~~[y-8]^2~~ = ~~\left( \frac{\sqrt{72}}{2} \right)^2\implies (x+7)^2~~ + ~~(y-8)^2~~ = ~~18$