The equation of a circle is $3x^{2} + 3y^{2} + 6x - 12y + 6 = 0$ . Find its radius

$\sqrt{3}$

$\sqrt{11}$

$\sqrt{6}$

Correct answer is B

The equation of a circle is given as $(x - a)^{2} + (y - b)^{2} = r^{2}$

Expanding this, we have $x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}$

Comparing with the given equation, $3x^{2} + 3y^{2} + 6x -12y + 6 = 0 \equiv x^{2} + y^{2} + 2x - 4y + 2 = 0$ (making the coefficients of $x^{2}$ and $y^{2}$ = 1 , we get that

$-2a = 2 \implies a = -1$

$2b = 4 \implies b = 2$

$r^{2} - a^{2} - b^{2} = -2$

$\therefore r^{2} - (-1)^{2} - (2)^{2} = -2 \implies r^{2} = -2 + 1 + 4 = 3$

$r = \sqrt{3}$

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