Question

C is the center of the circle with a diameter whose endpoints are the points A(-2, 1) and B(6, 9).

What is the equation of circle C?

Answer 1

well, the midpoint of segment AB is the center of the circle, so

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad B(\stackrel{x_2}{6}~,~\stackrel{y_2}{9}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 6 -2}{2}~~~ ,~~~ \cfrac{ 9 +1}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ 10 }{2} \right)\implies \stackrel{center}{(2~~,~~5)}`

now since we know AB is the diameter, so half that distance AB is its radius, so

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad B(\stackrel{x_2}{6}~,~\stackrel{y_2}{9})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2)`

`AB=√((~~6 - (-2)~~)^2 + (~~9 - 1~~)^2) \implies AB=√((6 +2)^2 + (9 -1)^2) \\\\\\ AB=√(( 8 )^2 + ( 8 )^2) \implies AB=√( 64 + 64 ) \implies AB=√( 128 ) \\\\\\ AB=8√(2)\hspace{5em}\stackrel{\textit{half that is the radius}}{\cfrac{8√(2)}{2}\implies 4√(2)\implies √(32)} \\\\[-0.35em] ~\dotfill`

`\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{2}{h}~~,~~\underset{5}{k})}\qquad \stackrel{radius}{\underset{√(32)}{r}} \\\\\\ (x-2)^2~~ + ~~(y-5)^2~~ = ~~(√(32))^2\implies {\large \begin{array}{llll} (x-2)^2~ + ~(y-5)^2~ = ~32 \end{array}}`