Find the equation of a circle with centre (-3, -8) and radius \(4\sqrt{6}\).
FURTHER MATHEMATICS
WAEC 2011
Find the equation of a circle with centre (-3, -8) and radius \(4\sqrt{6}\).
- A. \(x^{2} - y^{2} - 6x + 16y + 23 = 0\)
- B. \(x^{2} + y^{2} + 6x + 16y - 23 = 0\)
- C. \(x^{2} + y^{2} + 6x - 16y + 23 = 0\)
- D. \(x^{2} + y^{2} - 6x + 16y + 23 = 0\)
Correct Answer: B. \(x^{2} + y^{2} + 6x + 16y - 23 = 0\)
Explanation
Equation of a circle: \((x - a)^{2} + (y - b)^{2} = r^{2}\)
where (a, b) and r are the coordinates of the centre and radius respectively.
Given : \((a, b) = (-3, -8); r = 4\sqrt{6}\)
\((x - (-3))^{2} + (y - (-8))^{2} = (4\sqrt{6})^{2}\)
\(x^{2} + 6x + 9 + y^{2} + 16y + 64 = 96\)
\(x^{2} + y^{2} + 6x + 16y + 9 + 64 - 96 = 0\)
\(\implies x^{2} + y^{2} + 6x + 16y - 23 = 0\)
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