Question

Find an equation of the circle that has center −5, 6 and passes through −1, 1.

Answer 1

well, the distance from the center to a point on the circle is just its radius, so the distance from the center at (-5 , 6) to (-1 , 1) is just that, so

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-5}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{-1})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ r=√((~~-1 - (-5)~~)^2 + (~~-1 - 6~~)^2)\implies r=√((-1 +5)^2 + (-1 -6)^2) \\\\\\ r=√( (4)^2 + (-7)^2) \implies r=√( 16 + 49)\implies r=√( 65 )`

`~\dotfill\\\\ \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{-5}{h}~~,~~\underset{6}{k})}\qquad \stackrel{radius}{\underset{√(65)}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - (-5) ~~ )^2 ~~ + ~~ ( ~~ y-6 ~~ )^2~~ = ~~(√(65))^2\implies (x+5)^2 + (y-6)^2 = 65`