The equation of the circle,concentric with th | vedclass

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(C) The given circle is $x^2+y^2-6x-4y-12=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3$ and $f=-2$. The center of the circle is $(-g, -f

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$x^2+y^2-6x-4y+9=0$

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(C) The given circle is $x^2+y^2-6x-4y-12=0$. Comparing this with $x^2+y^2+2gx+2fy+c=0$,we get $g=-3$ and $f=-2$. The center of the circle is $(-g, -f) = (3, 2)$. Since the required circle is concentric,its center is also $(3, 2)$. Because the circle touches the $X$-axis,its radius $r$ is equal to the absolute value of the $y$-coordinate of its center,so $r = |2| = 2$. The equation of the circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. Substituting the values,we get $(x-3)^2 + (y-2)^2 = 2^2$. Expanding this,we get $(x^2-6x+9) + (y^2-4y+4) = 4$. Simplifying,we get $x^2+y^2-6x-4y+9 = 0$.

The equation of the circle,concentric with the circle $x^2+y^2-6x-4y-12=0$ and touching the $X$-axis is

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