Find the coordinates of the centre and the radius of the circles whose equations are
(a) $3 x^{2} + 3 y^{2} - 5 x - 6 y + 4 = 0$
(b) $4 x^{2} + 4 y^{2} - 16 x - 12 y + 21 = 0$ .

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Solution

$a) I n x^{2} + y^{2} + 2 g x + 2 f y + c = 0 C e n t r e i s (- g, - f) a n d r a d i u s = \sqrt{g^{2} + f^{2} - c} H e r e e q u a t i o n i s : \Rightarrow x^{2} + y^{2} - \frac{5 x}{3} - \frac{6 y}{3} + \frac{4}{3} = 0. ∴ C e n t r e \to (\frac{5}{6}, \frac{6}{6}) \to (\frac{5}{6}, 1) R a d i u s = \sqrt{\frac{25}{36} + 1 - \frac{4}{3}} = \sqrt{\frac{25 + 36 - 48}{36}} = \sqrt{\frac{13}{36}} . b) H e r e e q u a t i o n i s : x^{2} + y^{2} - \frac{16 x}{4} - \frac{12 y}{4} + \frac{21}{4} = 0. x^{2} + y^{2} - 4 x - 3 y + \frac{21}{4} = 0. ∴ C e n t r e \to (2, \frac{3}{2}) R a d i u s = \sqrt{4 + \frac{9}{4} - \frac{21}{4}} = \sqrt{4 - \frac{12}{4}} = 1 A n s .$

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