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.
A. 1
B. 2
C. 3
D. 4
Show Answer Show Explanation Correct Answer: C Explanation \(x^{2} + y^{2} - 2\alpha x + 4y - \alpha = 0\) r = 4 units; centre (\(\alpha\), -2). \((x - x_{1})^{2} + (y - y_{1})^{2} = r^{2}\) \((x - \alpha)^{2} + (y - (-2))^{2} = 4^{2}\) \(x^{2} - 2\alpha x + \alpha^{2} + y^{2} + 4y + 4 = 16\) \(x^{2} + y^{2} - 2\alpha x + 4y = 16 - \alpha^{2} - 4\) \(\therefore 12 - \alpha^{2} = \alpha \implies \alpha^{2} + \alpha - 12 = 0\) \(\alpha^{2} - 3\alpha + 4\alpha - 12 = 0\) \(\alpha(\alpha - 3) + 4(\alpha - 3) = 0\) \((\alpha + 4)(\alpha - 3) = 0 \implies +\alpha = 3\)