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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the centre and radius of the circle $2x^2+2y^2-8x-7y=0$

$\begin{array}{1 1} (A) center = (2, \large\frac{-7}{4}), radius = \large\frac{1}{4} \sqrt 113 \\ (B) center = (2, \large\frac{7}{4}), radius = \large\frac{1}{4} \sqrt 113 \\ (C) center = (2, \large\frac{-7}{4}), radius = \large\frac{1}{4} \sqrt 43 \\ (D) center = (2, \large\frac{7}{4}), radius = \large\frac{1}{4} \sqrt 43 \end{array} $

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1 Answer

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The general equation of a circle is given by $x^2+y^2+2gx+2fy+c$ where the center is given by $(-g,-f)$ and the radius by $r = \sqrt (g^2+f^2-c)$
Lets start by finding the center of the circle. From the equation, we can determine that $g=-2$ and $f=\large\frac{-7}{4}$$ \rightarrow $ center is $(2, \large\frac{7}{4}$$)$
The radius of the circle then is $r = \sqrt (g^2+f^2-c) = \sqrt ( (-2)^2 + (\large\frac{-7}{4})\normalsize^2)$ (We see from the equation that $c = 0$)
$\qquad r = \sqrt (4 + \large\frac{49}{16}$ $) = \large\frac{1}{4} $$\sqrt 113$
answered Mar 21, 2014 by balaji.thirumalai
 

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