The equation of a circle is (3x^{2} + 3y^{2} + 6x – 12y + 6 = 0). Find its radius
-
A.
1 -
B.
(sqrt{3}) -
C.
(sqrt{11}) -
D.
(sqrt{6})
Correct Answer: Option B
Explanation
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})