Question

A diameter of a circle has its endpoints at (-2,-1) & (4,7). What is the equation of the circle.

Answer 1

Check the picture below.

so the center of the circle is the midpoint of that diameter, and its radius is half its length.

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{7}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 4 -2}{2}~~~ ,~~~ \cfrac{ 7 -1}{2} \right) \implies \left(\cfrac{ 2 }{2}~~~ ,~~~ \cfrac{ 6 }{2} \right)\implies \stackrel{ center }{(1~~,~~3)} \\\\[-0.35em] ~\dotfill`

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ r(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{7})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ \stackrel{ radius }{r}=√((~~4 - (-2)~~)^2 + (~~7 - (-1)~~)^2) \implies r=√((4 +2)^2 + (7 +1)^2) \\\\\\ r=√(( 6 )^2 + ( 8 )^2) \implies r=√( 36 + 64 ) \implies r=√( 100 )\implies r=10 \\\\[-0.35em] ~\dotfill`

`\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{1}{h}~~,~~\underset{3}{k})}\qquad \stackrel{radius}{\underset{10}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - 1 ~~ )^2 ~~ + ~~ ( ~~ y-3 ~~ )^2~~ = ~~10^2\implies {\large \begin{array}{llll} (x-1)^2+(y-3)=100 \end{array}}`