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(D) The lines passing through $(3,2)$ and parallel to the coordinate axes are $x=3$ and $y=2$. Since the circle $C$ of radius $r=1$ touches these line

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$4$

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(D) The lines passing through $(3,2)$ and parallel to the coordinate axes are $x=3$ and $y=2$. Since the circle $C$ of radius $r=1$ touches these lines and is closer to the origin,its center $(h,k)$ must be at a distance of $1$ from these lines such that $h Thus,the center is $(3-1, 2-1) = (2,1)$.The equation of the circle is $(x-2)^2 + (y-1)^2 = 1^2$.The distance from the center $C(2,1)$ to the point $Q(5,5)$ is $CQ = \sqrt{(5-2)^2 + (5-1)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = 5$.The shortest distance from the circle to the point $Q$ is $CQ - r = 5 - 1 = 4$.

Let a circle $C$ of radius $1$ and closer to the origin be such that the lines passing through the point $(3,2)$ and parallel to the coordinate axes touch it. Then the shortest distance of the circle $C$ from the point $(5,5)$ is:

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