Find a positive value of \(\alpha\) if the coordinate of the centre of a circle x\(^2\) + y\(^2\) – 2\(\alpha\)x + 4y – \(\alpha\) = 0 is (\(\alpha\), -2) and the radius is 4 units.
- 1
- 2
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3
- 4
The correct answer is: C
Explanation
\(x^{2} + y^{2} - 2\alpha x + 4y - \alpha = 0\)
r = 4 units; centre (\(\alpha\), -2).
\((x - x_{1})^{2} + (y - y_{1})^{2} = r^{2}\)
\((x - \alpha)^{2} + (y - (-2))^{2} = 4^{2}\)
\(x^{2} - 2\alpha x + \alpha^{2} + y^{2} + 4y + 4 = 16\)
\(x^{2} + y^{2} - 2\alpha x + 4y = 16 - \alpha^{2} - 4\)
\(\therefore 12 - \alpha^{2} = \alpha \implies \alpha^{2} + \alpha - 12 = 0\)
\(\alpha^{2} - 3\alpha + 4\alpha - 12 = 0\)
\(\alpha(\alpha - 3) + 4(\alpha - 3) = 0\)
\((\alpha + 4)(\alpha - 3) = 0 \implies +\alpha = 3\)