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(B) The equation of a circle concentric with the given circle $x^2 + y^2 - 4x - 6y - 3 = 0$ is of the form $x^2 + y^2 - 4x - 6y + k = 0$.The center of

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

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(B) The equation of a circle concentric with the given circle $x^2 + y^2 - 4x - 6y - 3 = 0$ is of the form $x^2 + y^2 - 4x - 6y + k = 0$.The center of this circle is $(2, 3)$.Since the circle touches the $y$-axis $(x = 0)$,the radius of the circle is equal to the absolute value of the $x$-coordinate of the center.Radius $r = |2| = 2$.The formula for the radius of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $\sqrt{g^2 + f^2 - c}$.Here,$g = -2$,$f = -3$,and $c = k$.So,$\sqrt{(-2)^2 + (-3)^2 - k} = 2$.Squaring both sides,we get $4 + 9 - k = 4$.$13 - k = 4 \Rightarrow k = 9$.Substituting $k = 9$ into the equation,we get $x^2 + y^2 - 4x - 6y + 9 = 0$.

The equation of the circle concentric with the circle $x^2 + y^2 - 4x - 6y - 3 = 0$ and touching the $y$-axis is:

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