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Find the radius of the circle \(2x^2 + 2y^2 - 4x + 5y + 1...

Further Mathematics
WAEC 2023

Find the radius of the circle \(2x^2 + 2y^2 - 4x + 5y + 1 = 0\)

  • A. \(\frac{\sqrt33}{4}\)
  • B. \(\frac{\sqrt5}{6}\)
  • C. \(\frac{5}{6}\)
  • D. \(\frac{33}{4}\)
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Correct Answer: Option A
Explanation

Standard Form equation of a circle (Center-Radius Form): \((x − a)^2 + (y − b)^2 = r^2\)

Where "a" and "b" are the coordinates of the center and "r" is the radius of the circle

\(2x^2+2y^2-4x+5y+1=0\)

Divide through by 2

= \(x^2+y^2-2x+\frac{5}{2}y+\frac{1}{2}= 0\)

=\(x^2-2x+y^2+\frac{5}{ 2}y=-\frac{1}{ 2}\)

=\(x^2-2x+1^2+y^2+\frac{5}{ 2}y+(\frac{5}{ 4})^2-1-\frac{25}{16}=-\frac{1}{2}\)

=\((x-1)^2+(y+\frac{5}{4})^2=-\frac{1}{2}+1+\frac{25}{16}\)

=\((x-1)^2+(y-(-\frac{5}{4}))^2=\frac{33}{16}\)

=\((x-1)^2+(y-(-\frac{5}{4}))^2=(\frac{\sqrt33}{4})^2\)

\(\therefore a = 1, b = - \frac{5}{4} and\) \(r \frac{\sqrt33}{4} (answer)\)


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WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709
WAEC offline past questions - with all answers and explanations in one app - Download for free
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
WAEC May/June 2024 - Practice for Objective & Theory - From 1988 till date, download app now - 99995
Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
WAEC offline past questions - with all answers and explanations in one app - Download for free
WAEC Past Questions, Objective & Theory, Study 100% offline, Download app now - 24709