Question

A circle has a diameter with endpoints at (-2,5) and (4,1). What is the equation of the circle? Select all that apply.

Answer 1

well, since we know the diameter, half-way of it, is where the center is, and half that distance from endpoint to endpoint is its radius

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 4 -2}{2}~~~ ,~~~ \cfrac{ 1 +5}{2} \right) \implies \left(\cfrac{ 2 }{2}~~~ ,~~~ \cfrac{ 6 }{2} \right)\implies \stackrel{ center }{(1~~,~~3)} \\\\[-0.35em] ~\dotfill`

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{1})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ d=√((~~4 - (-2)~~)^2 + (~~1 - 5~~)^2) \implies d=√((4 +2)^2 + (1 -5)^2) \\\\\\ d=√(( 6 )^2 + ( -4 )^2) \implies d=√( 36 + 16 ) \implies d=√( 52 ) \\\\[-0.35em] ~\dotfill\\`

`\stackrel{\textit{half of that diameter is its radius}}{r=\cfrac{√(52)}{2}\implies r^2=\cfrac{52}{4}}\implies r^2=13 \\\\[-0.35em] \rule{34em}{0.25pt}`

`\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{(√(52))/(2)}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - 1 ~~ )^2 ~~ + ~~ ( ~~ y-3 ~~ )^2~~ = ~~\left( (√(52))/(2) \right)^2\implies \boxed{(x-1)^2 + (y-3)^2=13} \\\\\\ (x^2-2x+1)+(y^2-6y+9)=13\implies \boxed{x^2+y^2-2x-6y=3}`