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(A) Given circle: $x^2+y^2-6x+12y+15=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=6, c=15$. Centre is $(-g, -f) = (3, -6)$. Radius $r = \s

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

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(A) Given circle: $x^2+y^2-6x+12y+15=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=6, c=15$. Centre is $(-g, -f) = (3, -6)$. Radius $r = \sqrt{g^2+f^2-c} = \sqrt{9+36-15} = \sqrt{30}$. Area of the given circle $A = \pi r^2 = 30\pi$. The required circle is concentric,so it has the same centre $(3, -6)$. Let the radius of the required circle be $R$. Given that the area of the required circle is twice the area of the given circle: $\pi R^2 = 2 	imes 30\pi = 60\pi$. $R^2 = 60$. The equation of the required circle is $(x-3)^2 + (y+6)^2 = 60$. $x^2 - 6x + 9 + y^2 + 12y + 36 = 60$. $x^2 + y^2 - 6x + 12y + 45 - 60 = 0$. $x^2 + y^2 - 6x + 12y - 15 = 0$.

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

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