Question

A circle has (-1, -1) and (-25,-11) as endpoints of a diameter.

Find the center and radius of the circle.
Write the standard equation of the circle.

Answer 1

well, since we know the diameter points, half-way in between is the center

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{-25}~,~\stackrel{y_2}{-11}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -25 -1}{2}~~~ ,~~~ \cfrac{ -11 -1}{2} \right)\implies \left(\cfrac{-26}{2}~~,~~\cfrac{-12}{2} \right)\implies (-13~~,~~-6)`

and its radius will be half the length of the diameter

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{-25}~,~\stackrel{y_2}{-11})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ d=√([-25 - (-1)]^2 + [-11 - (-1)]^2)\implies d=√((-25+1)^2+(-11+1)^2) \\\\\\ d=√((-24)^2+(-10)^2)\implies d=√(676)\implies d=26~\hfill \stackrel{half~that}{r=13}`

`\rule{34em}{0.25pt}\\\\ \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-13}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{13}{ r} \\\\[-0.35em] ~\dotfill\\\\\ [x-(-13)]^2~~ + ~~[y-(-6)]^2~~ = ~~13^2\implies (x+13)^2~~ + ~~(y+6)^2~~ = ~~169`