The equation of the circle concentric with th | vedclass

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(A) The given circle is $x^2+y^2-6x+12y+15=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=6, c=15$. The radius $r_1$ is $\sqrt{g^2+f^2-c} =

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

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(A) The given circle is $x^2+y^2-6x+12y+15=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3, f=6, c=15$. The radius $r_1$ is $\sqrt{g^2+f^2-c} = \sqrt{(-3)^2+6^2-15} = \sqrt{9+36-15} = \sqrt{30}$. The area of the given circle is $A_1 = \pi r_1^2 = 30\pi$. Let the concentric circle be $x^2+y^2-6x+12y+k=0$. Its radius $r_2$ satisfies $r_2^2 = g^2+f^2-k = 45-k$. According to the problem,the area of the new circle is $A_2 = 2A_1 = 60\pi$. Thus,$\pi r_2^2 = 60\pi$,which implies $r_2^2 = 60$. Substituting $r_2^2 = 45-k$,we get $45-k = 60$,so $k = -15$. Therefore,the equation of the circle is $x^2+y^2-6x+12y-15=0$.

The equation of the circle concentric with the circle $x^2+y^2-6x+12y+15=0$ and having double its area is:

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