Mathematics, 21.08.2019 02:20 sydneykated
The center of a circle is (3, -1). one point on the circle is (6,2). find the equation of the circle
Answers: 1
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{3}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{[6-3]^2+[2-(-1)]^2}\implies r=\sqrt{(6-3)^2+(2+1)^2} \\\\\\ r=\sqrt{3^2+3^2}\implies r=\sqrt{18} \\\\[-0.35em] ~\dotfill](/tpl/images/0183/4472/1d56a.png)
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{3}{ h},\stackrel{-1}{ k})\qquad \qquad radius=\stackrel{\sqrt{18}}{ r}\\[2em] [x-3]^2+[y-(-1)]^2=(\sqrt{18})^2\implies (x-3)^2+(y+1)^2=18](/tpl/images/0183/4472/1c21a.png)
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The center of a circle is (3, -1). one point on the circle is (6,2). find the equation of the circle...
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