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

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

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(A) The given equation of the circle is $x^2+y^2-6x-4y-12=0$. Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-3$ and $f=-2$. The center of the circle is $(-g, -f) = (3, 2)$. $A$ circle concentric with the given circle will have the same center $(3, 2)$. Let the equation of the required circle be $(x-3)^2+(y-2)^2=r^2$. Since the circle touches the $Y$-axis,the radius $r$ is equal to the absolute value of the $x$-coordinate of the center,so $r = |3| = 3$. Substituting $r=3$ into the equation: $(x-3)^2+(y-2)^2=3^2$. Expanding this,we get $x^2-6x+9+y^2-4y+4=9$. Simplifying,we get $x^2+y^2-6x-4y+4=0$.

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

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