Question

What is the equation of the circle with center (11, 6) that passes through the point (17, 12)?

Answer 1

well, the distance from the center of a circle to any point "on" the circle is simply just its radius, thus

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{center}{(\stackrel{x_1}{11}~,~\stackrel{y_1}{6})}\qquad (\stackrel{x_2}{17}~,~\stackrel{y_2}{12})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{(17-11)^2+(12-6)^2}\implies r = \sqrt{36+36}\implies r=\sqrt{72} \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{11}{ h},\stackrel{6}{ k})\qquad \qquad radius=\stackrel{\sqrt{72}}{ r} \\\\\\ (x-11)^2+(y-6)^2=(\sqrt{72})^2\implies (x-11)^2+(y-6)^2=72$