The equation of a circle is \(3x^{2} + 3y^{2} + 6x - 12y + 6 = 0\). Find its radius
Answer Details
To find the radius of the given circle, we need to rewrite the equation in standard form, which is of the form \((x - a)^2 + (y - b)^2 = r^2\), where \((a, b)\) is the center of the circle and \(r\) is its radius.
Completing the square for the given equation, we have:
\begin{align*}
3x^{2} + 3y^{2} + 6x - 12y + 6 &= 0\\
3(x^{2} + 2x) + 3(y^{2} - 4y) &= -6\\
3(x^{2} + 2x + 1) + 3(y^{2} - 4y + 4) &= -6 + 3 + 12\\
3(x + 1)^{2} + 3(y - 2)^{2} &= 9
\end{align*}
Dividing both sides by 3, we have:
\[(x + 1)^{2} + (y - 2)^{2} = 3\]
Comparing this with the standard form, we can see that the center of the circle is \((-1, 2)\), and the radius is \(\sqrt{3}\).
Therefore, the answer is \boxed{\sqrt{3}}.