The equation of a circle is \(3x^{2} + 3y^{2} + 6x - 12y + 6 = 0\). Find its radius
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
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
\(-2a = 2 \implies a = -1\)
\(2b = 4 \implies b = 2\)
\(r^{2} - a^{2} - b^{2} = -2\)
\(\therefore r^{2} - (-1)^{2} - (2)^{2} = -2 \implies r^{2} = -2 + 1 + 4 = 3\)
\(r = \sqrt{3}\)
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
\(-2a = 2 \implies a = -1\)
\(2b = 4 \implies b = 2\)
\(r^{2} - a^{2} - b^{2} = -2\)
\(\therefore r^{2} - (-1)^{2} - (2)^{2} = -2 \implies r^{2} = -2 + 1 + 4 = 3\)
\(r = \sqrt{3}\)