Respuesta :

[tex]The\ equation\ of\ the\ circle:(x-a)^2+(y-b)^2=r^2\\\\where\ (a;\ b)\ -the\ coordinates\ of\ the\ center;\ r-the\ radius\\-----------------------------\\(-1;-2)-the\ center\ of\ the\ circle\\r=5\\\\The\ equation:(x+1)^2+(y+2)^2=5^5\\\\Put\ x=2\ to\ the\ equation:\\\\(2+1)^2+(y+2)^2=25\\3^2+(y+2)^2=25\\9+(y+2)^2=25\\(y+2)^2=25-9\\(y+2)^2=16\iff y+1=\pm\sqrt{16}\\\\y+1=-4\ or\ y+1=4\ \ \ \ \ |subtract\ 1\ from\ both \sides\\y=-5\ or\ y=3\\\\Answer:(2;-5)\ and\ (2;\ 3).[/tex]

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