Question

Write the equation of the circle with the segment whose endpoints are (-2,5) and (4,7) as the diameter. (Hint.... the center of a circle is the midpoint of the diameter.)

Answer 1

yeap, the midpoint is the center of it and half the distance between those two points is the radius.

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\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 +5}{2} \right) \implies \left(\cfrac{ 2 }{2}~~~ ,~~~ \cfrac{ 12 }{2} \right)\implies \stackrel{ center }{(1~~,~~6)} \\\\[-0.35em] ~\dotfill`

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{7})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ \stackrel{ diameter }{d}=√((~~4 - (-2)~~)^2 + (~~7 - 5~~)^2) \implies d=√((4 +2)^2 + (7 -5)^2) \\\\\\ d=√(( 6 )^2 + ( 2 )^2) \implies d=√( 36 + 4 ) \implies d=√( 40 )~\hfill \stackrel{radius}{\cfrac{√(40)}{2}} \\\\[-0.35em] ~\dotfill`

`\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{1}{h}~~,~~\underset{6}{k})}\qquad \stackrel{radius}{\underset{(√(40))/(2)}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - 1 ~~ )^2 ~~ + ~~ ( ~~ y-6 ~~ )^2~~ = ~~\left( (√(40))/(2) \right)^2\implies \boxed{(x-1)^2+(y-6)^2=10}`