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(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. $(i)$Since the circle touches the $X$-axis,the radius is equal to the absolute value of t

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$(2,-1)$

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(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. $(i)$Since the circle touches the $X$-axis,the radius is equal to the absolute value of the $y$-coordinate of the centre,so $r^2 = f^2$. Thus,$g^2+f^2-c = f^2$,which implies $c = g^2$. $(ii)$The circle passes through $(2,0)$,so $4+0+4g+0+c=0$,which gives $c = -4g-4$. $(iii)$Equating $(ii)$ and $(iii)$,we get $g^2 = -4g-4$,so $g^2+4g+4=0$,which means $(g+2)^2=0$,so $g=-2$. Substituting $g=-2$ into $(ii)$,we get $c=(-2)^2=4$.The circle $x^2+y^2-2x-2y-2=0$ has $g_1=-1, f_1=-1, c_1=-2$. The condition for orthogonality is $2g_1g_2+2f_1f_2 = c_1+c_2$.Substituting the values: $2(-1)(-2) + 2(-1)(f) = -2 + 4$,which simplifies to $4 - 2f = 2$,so $2f = 2$,which gives $f=1$.The centre of the circle is $(-g, -f) = (-(-2), -1) = (2, -1)$.

The centre of the circle which intersects the circle $x^2+y^2-2x-2y-2=0$ orthogonally,passes through the point $(2,0)$,and touches the $X$-axis is:

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