Question

Write the equation of the circle whose endpoints of a diameter are (2,-5) and (8,-1)

Answer 1

if the endpoints are there, that means that segment with those endpoints is the diameter of the circle, and that also means that the midpoint of that diameter is the center of the circle.

it also means that the distance from the midpoint to either endpoint, is the radius of the circle.

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{-5})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{8+2}{2}~~,~~\cfrac{-1-5}{2} \right)\implies (5,-3)\impliedby center \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{midpoint}{(\stackrel{x_1}{5}~,~\stackrel{y_1}{-3})}\qquad \stackrel{endpoint}{(\stackrel{x_2}{8}~,~\stackrel{y_2}{-1})}\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{[8-5]^2+[-1-(-3)]^2}\implies r=\sqrt{(8-5)^2+(-1+3)^2} \\\\\\ r=\sqrt{3^2+2^2}\implies r=\sqrt{13} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{5}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{\sqrt{13}}{ r} \\[2em] [x-5]^2+[y-(-3)]^2=(\sqrt{13})^2\implies \boxed{(x-5)^2+(y+3)^2=13}$