Question

Find the equation for the circle with a diameter whose endpoints are (-4,4) and (5,3).

Answer 1

well, we know the diameter endpoints, so hmmm the center of the circle will be half-way of the diameter, or namely its midpoint and its radius is half the length of the diameter, let's get both.

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 5 -4}{2}~~~ ,~~~ \cfrac{ 3 +4}{2} \right) \implies \stackrel{ center }{\left(\cfrac{ 1 }{2}~~~ ,~~~ \cfrac{ 7 }{2} \right)} \\\\[-0.35em] ~\dotfill`

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{3})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ d=√((~~5 - (-4)~~)^2 + (~~3 - 4~~)^2)\implies d=√((5 +4)^2 + (3 -4)^2) \\\\\\ d=√( (9)^2 + (-1)^2) \implies d=√( 81 + 1)\implies d=√( 82 )~\hfill~\stackrel{radius}{\cfrac{√(82)}{2}} \\\\[-0.35em] ~\dotfill`

`\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{h}{(1)/(2)}~~,~~\underset{k}{(7)/(2)})} \qquad \stackrel{radius}{\underset{r}{(√(82))/(2)}} \\\\[-0.35em] ~\dotfill\\\\ ( x - (1)/(2) )^2 ~ + ~ ( y-(7)/(2) )^2~~ = ~~\left( (√(82))/(2) \right)^2 \implies ( ~ x - (1)/(2) ~ )^2 ~~ + ~~ ( ~ y-(7)/(2) ~ )^2 = (41)/(2)`