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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. \(x^{2} + y^{2} - 6x + 16y + 23 = 0\)
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Correct Answer: Option B
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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Project Topics and Materials, Post-UTME past questions - Download now or upload yours to get paid
NECO June/July 2024 - Get offline past questions & answers - Download objective & theory, all in one app 48789
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
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Join your school's WhatsApp group
Project Topics and Materials, Post-UTME past questions - Download now or upload yours to get paid
NECO June/July 2024 - Get offline past questions & answers - Download objective & theory, all in one app 48789
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts