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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equation of the circle with centre (–2,3) and radius 4

$\begin {array} {1 1} (A)\;x^2-y^2-4x+6y-3=0 & \quad (B)\;x^2+y^2-4x+6y-3=0 \\ (C)\;x^2+y^2+4x-6y-3=0 & \quad (D)\;x^2+y^2-4x-6y+3=0 \end {array}$

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  • Equation of a circle with centre (h, k) and radius r is given as : $(x-h)^2+(y-k)^2=r^2$
The coordinate of the centre is (-2, 3) and radius is 4.
Hence the equation of the circle is $(x+2)^2+(y-3)^2=4^2$
On expanding we get,
$x^2+4x+4+y^2-6y+9=16$
(i.e) $x^2+y^2+4x-6y-3=0$
is the required equation of the circle.
answered Jul 11, 2014 by thanvigandhi_1
 

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