Question

Find the equation of the circle passing through the origin and with the center (3,5)

Answer 1

$Equation\ of\ circle\\\\\ r^2=(x-a)^2+(y-b)^2\\\\ a,b- \ coordinates\ of\ center\ of\ circle\\\\\ (a,b)=(3,5)\\\\ r^2=(x-3)^2+(y-5)^2\\\\\ Substituting\ point\ (0,0)-origin\\\\ r^2=(0-3)^2+(0-5)^2\\\\ r^2=(-3)^2+(-5)^2\\\\ r^2=9+25=36\\ r=6\\\\ Answer:\ \ 36=(x-3)^2+(y-5)^2$