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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

Find the centre and radius of the circle $x^2+y^2-2x-4y=9$.

$\begin{array}{1 1}(a)\;(1,2),\sqrt{14}\\(b)\;(-1,-2),\sqrt{13}\\(c)\;(1,4),\sqrt{11}\\(d)\;(2,4),\sqrt{15}\end{array}$

1 Answer

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  • Centre of the circle $x^2+y^2+2gx+2fy+c=0$ is $(-g,-f)$ and radius $=\sqrt {g^2+f^2-c}$
Given equation of the circle is $x^2+y^2-2x-4y=9$
Comparing this equation with the general equation of the circle, we have
$2g=-2,\:2f=-4\:\;and\:\:c=-9$
$\therefore \:Centre=(-g,-f)=(1,2)$
$Radius=\sqrt {g^2+f^2-c}=\sqrt{1^2+2^2+9}=\sqrt {14}$
Radius=$\sqrt{14}$
Hence (a) is the correct answer.
answered Feb 11, 2014 by sreemathi.v
edited Mar 24, 2014 by rvidyagovindarajan_1
 

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