WAEC - Further Mathematics (2021 - No. 34)
Find the radius of the circle 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0.
\(\frac{17}{4}\)
\(\frac{17}{2}\)
\(\frac{17}{\sqrt{2}}\)
\(\frac{\sqrt{17}}{2}\)
Explanation
2x\(^3\) - 4x + 2y\(^2\) - 6y - 2 = 0
Divide through by 2: x\(^2\) - 2x + y\(^2\) -3y -1 = 0
x\(^2\) -2x + y\(^2\) - 3y = 1
x\(^2\) -2x + 1 + y\(^2\) - 3y + \(\frac{9}{4}\)
= 1+ 1 + \(\frac{9}{4}\)
= (x- 1)\(^2\) (y - 3/2)\(^2\)
= \(\sqrt{\frac{17}{4}}\)
r = \(\frac{\sqrt{17}}{2}\)
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