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(D) The given equation of the circle is $x^2+y^2-2x+ky+4=0$  $(i)$.The center of the circle $(i)$ is $C \equiv (1, -k/2)$.Since the circle $S$ is conc

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$\frac{1}{2}\sqrt{137}$

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(D) The given equation of the circle is $x^2+y^2-2x+ky+4=0$  $(i)$.The center of the circle $(i)$ is $C \equiv (1, -k/2)$.Since the circle $S$ is concentric with circle $(i)$,the equation of $S$ is $(x-1)^2 + (y+k/2)^2 = r^2$ $(ii)$.The center $(1, -k/2)$ lies on the diameter line $3x-2y+4=0$.Substituting the center coordinates into the line equation: $3(1) - 2(-k/2) + 4 = 0$ $\Rightarrow 3 + k + 4 = 0$ $\Rightarrow k = -7$.Substituting $k = -7$ into equation $(ii)$,we get $(x-1)^2 + (y-7/2)^2 = r^2$ $(iii)$.Since the circle $S$ passes through the point $(3, -2)$,we substitute these coordinates into equation $(iii)$:$(3-1)^2 + (-2-7/2)^2 = r^2$$2^2 + (-11/2)^2 = r^2$$4 + 121/4 = r^2$$r^2 = (16+121)/4 = 137/4$.Therefore,the radius $r = \frac{\sqrt{137}}{2} = \frac{1}{2}\sqrt{137}$.

Let the circle $S$ which is concentric with the circle $x^2+y^2-2x+ky+4=0$ pass through the point $(3,-2)$. If one of the diameters of $S$ lies along the line $3x-2y+4=0$,then the radius of the circle $S$ is

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