Find the centre and radius of the circle $2 x^{2} + 2 y^{2} = 3 x - 5 y + 7$ .

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Solution

$\begin{matrix} T o f i n d t h e c e n t r e a n d r a d i u s o f c e n t r e 2 x^{2} + 2 y^{2} = 3 x - 5 y + 7 2 x^{2} + 2 y^{2} - 3 x + 5 y = 7 x^{2} + y^{2} - \frac{3}{2} x + \frac{5}{2} y = \frac{7}{2} (\begin{matrix} d e v i d e d b y 2 \end{matrix}) x^{2} - \frac{3}{2} x + y^{2} + \frac{5}{2} y = \frac{7}{2} x^{2} - \frac{3}{2} x + \frac{9}{16} + y^{2} + \frac{5}{2} y + \frac{25}{16} = \frac{7}{2} + \frac{9}{16} + \frac{25}{16} {(x - \frac{3}{4})}^{2} + {(y + \frac{5}{4})}^{2} = \frac{7}{2} + \frac{9}{16} + \frac{25}{16} = \frac{7}{2} + \frac{34}{16} = \frac{90}{16} \Rightarrow {(x - \frac{3}{4})}^{2} + {(y + \frac{5}{4})}^{2} = \frac{90}{16} C o m p a r i n g w i t h g e n e r a l e q u a t i o n h = \frac{3}{4}, k = \frac{- 5}{4}, r = \sqrt{\frac{9}{16}} (h, k) = (\frac{3}{4}, \frac{- 5}{4}) r = \frac{\sqrt{90}}{4} A n s . \end{matrix}$

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