Question

find the center of the circle defined by the equation 3x^2+3y^2+42x+42y+186=0

Answer 1

Complete the square in terms of \(x\) and \(y\).
\[\begin{align*}
3x^2+42x+3y^2+42y+186&=0\\
x^2+14x+y^2+14y+62&=0\\
x^2+14x+49+y^2+14y+49-36&=0\\
(x+7)^2+(y+7)^2&=36
\end{align*}\]