Answer :

[tex]Center\ of\ the\ circle\ is\ the\ midpoint\ AB\ where\ A(-2;\ 4)\ \&\ B(4;\ 2)\\\\center:O\left(\frac{-2+4}{2};\ \frac{4+2}{2}\right)\to O(1;\ 3)\\\\Radius\ of\ the\ circle\ is\ distance\ O\ to\ A:\\\\r=|OA|=\sqrt{(-2-1)^2+(4-3)^2}=\sqrt{9+1}=\sqrt{10}\\\\Equation\ of\ a\ circle:\\\\(x-a)^2+(y-b)^2=r^2\ whete\ center\ is\ (a;\ b)\ and\ radius\ is\ r.\\\\Equation\ of\ this\ circle:\\\\(x-1)^2+(y-3)^2=(\sqrt{10})^2\\\\(x-1)^2+(y-3)^2=10[/tex]

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