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The equation of a circle is \(x^{2} + y^{2} - 8x + 9y + 15...

Further Mathematics
WAEC 2013

The equation of a circle is \(x^{2} + y^{2} - 8x + 9y + 15 = 0\). Find its radius.

  • A. 5
  • B. \(\frac{1}{2}\sqrt{15}\)
  • C. \(\frac{1}{2}\sqrt{85}\)
  • D. \(\sqrt{85}\)
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Correct Answer: Option C
Explanation

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

Expanding, we have: \(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)

\(x^{2} + y^{2} - 2ax - 2by + a^{2} + b^{2} = r^{2}\)

Comparing with the equation, \(x^{2} + y^{2} - 8x + 9y = -15\), we have

\(2a = 8; 2b = -9; r^{2} - a^{2} - b^{2} = -15\)

\(a = 4; b = \frac{-9}{2}\)

\(\therefore  r^{2} = -15 + 4^{2} + (\frac{-9}{2})^{2}\)

= \(-15 + 16 + \frac{81}{4} = \frac{85}{4}\)

\(r = \sqrt{\frac{85}{4} = \frac{1}{2}\sqrt{85}\)


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WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
Join your school's WhatsApp group
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
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
NECO June/July 2024 - Get offline past questions & answers - Download objective & theory, all in one app 48789
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WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709