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(B) Let the equation of the required circle be $(x-h)^2+(y-k)^2=r^2$. Since it touches the $Y$-axis,the radius $r = |h|$. Thus,the equation is $(x-h)^

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

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(B) Let the equation of the required circle be $(x-h)^2+(y-k)^2=r^2$. Since it touches the $Y$-axis,the radius $r = |h|$. Thus,the equation is $(x-h)^2+(y-k)^2=h^2$,which simplifies to $x^2+y^2-2hx-2ky+k^2=0$. Since it passes through $(3,0)$,we have $(3-h)^2+(0-k)^2=h^2$,which simplifies to $9-6h+k^2=0$,or $k^2=6h-9$ ... $(i)$. The circle $x^2+y^2-6x+4y-3=0$ has $g_2=-3, f_2=2, c_2=-3$. The required circle has $g_1=-h, f_1=-k, c_1=k^2$. For orthogonal intersection,$2(g_1g_2+f_1f_2) = c_1+c_2$. Substituting the values: $2((-h)(-3)+(-k)(2)) = k^2-3$,which gives $6h-4k = k^2-3$. Substituting $k^2=6h-9$ from $(i)$ into this equation: $6h-4k = (6h-9)-3$,which simplifies to $-4k = -12$,so $k=3$. Substituting $k=3$ into $(i)$: $9-6h+9=0$,so $6h=18$,$h=3$. The center is $(3,3)$ and the radius is $3$. The equation is $(x-3)^2+(y-3)^2=3^2$,which simplifies to $x^2+y^2-6x-6y+9=0$.

Find the equation of a circle which cuts the circle $x^2+y^2-6x+4y-3=0$ orthogonally,while passing through $(3,0)$ and touching the $Y$-axis.

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