Question

Write the standard form of an equation of the circle with endpoints of a diameter (-3, 3) and (-1, -1

Answer 1

well, we know the endpoints for its diameter, so the midpoint of that will be where the center is at, and half the distance between those two endppoints is its radius, let's get those two.

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

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-3}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{-1})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ d=√((~~-1 - (-3)~~)^2 + (~~-1 - 3~~)^2)\implies d=√((-1 +3)^2 + (-1 -3)^2) \\\\\\ d=√( (2)^2 + (-4)^2) \implies d=√( 4 + 16)\implies d=√( 20 )~\hfill~\stackrel{radius}{\cfrac{√(20)}{2}} \\\\[-0.35em] ~\dotfill`

`\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{h}{-2}~~,~~\underset{k}{1})} \qquad \stackrel{radius}{\underset{r}{\cfrac{√(20)}{2}}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ( ~~ x - (-2) ~~ )^2 ~~ + ~~ ( ~~ y-1 ~~ )^2~~ = ~~\left( \cfrac{√(20)}{2} \right)^2 \\\\\\ (x +2)^2 + (y -1)^2 = \cfrac{20}{4}\implies \boxed{(x +2)^2 + (y -1)^2 =5}`