Mathematics | TS EAMCET 2019 | Discussion Page

The equation of a circle concentric with the circle $x^{2}+y^{2}-6x+12y+15=0$ and having area that is twice the area of the given circle is

$x^{2}+y^{2}-6x+12y-15=0$

$x^{2}+y^{2}-6x+12y-30=0$

$x^{2}+y^{2}-6x+12y-60=0$

$x^{2}+y^{2}-6x+12y+15=0$

Option A

Given,

$x^{2}+y^{2}-6x+12y+15=0$

centre (3,-6)

$r= \sqrt{9+36-15}= \sqrt{30}$

Area = $\pi (\sqrt{30})^{2}=30 \pi$

Required equation of circle has twice the area of circle

$\therefore$ $A'= 60 \pi$

$\pi R^{2}=60 \pi , R^{2}=60$

$\therefore$ Equation of circle concentric with given circle is

$(x-3)^{2}+(y+6)^{2}=60$

$x^{2}+y^{2}-6x+12y-15=0$

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