Question

Write the standard equation of a circle with center (-2,5) and the point (4,13) on the circle ​

Answer 1

since the point (4,13) is on the circle, then the distance from the center to it, is the radius of the circle.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{13})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{[4-(-2)]^2+[13-5]^2}\implies r=\sqrt{(4+2)^2+(13-5)^2} \\\\\\ r=\sqrt{36+64}\implies r=\sqrt{100}\implies r=10 \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-2}{ h},\stackrel{5}{ k})\qquad \qquad radius=\stackrel{10}{ r}\\[2em] [x-(-2)]^2+[y-5]^2=10^2\implies (x+2)^2+(y-5)^2=100$