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Find the equation of the circle with centre (–a, –b) and radius $ \sqrt{a^2-b^2}$

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

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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 coordinates of the centre are (-a, -b) and the radius is $ \sqrt{a^2-b^2}$
$ \therefore $ The equation of the circle is
$(x+a)^2+(y+b)^2=( \sqrt{a^2-b^2} )^2$
On expanding we get,
On rearranging we get,
is the required equation of the circle.
answered Jul 11, 2014 by thanvigandhi_1

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