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(C) Let the equation of the circle $S$ be $x^2+y^2+2gx+2fy+c=0$. Since the circle passes through the origin $(0,0)$,we have $c=0$. The centre of the c

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$2\sqrt{2}$

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(C) Let the equation of the circle $S$ be $x^2+y^2+2gx+2fy+c=0$. Since the circle passes through the origin $(0,0)$,we have $c=0$. The centre of the circle is $(-g, -f)$. Since the centre lies on the line $x-y=0$,we have $-g - (-f) = 0$,which implies $g=f$. Thus,the equation of the circle $S$ is $x^2+y^2+2gx+2gy=0$. Two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ cut orthogonally if $2g_1g_2 + 2f_1f_2 = c_1+c_2$. Here,$g_1=g, f_1=g, c_1=0$ and $g_2=-2, f_2=-3, c_2=10$. Substituting these values: $2(g)(-2) + 2(g)(-3) = 0 + 10$. $-4g - 6g = 10$ $\Rightarrow -10g = 10$ $\Rightarrow g = -1$. The radius of circle $S$ is $r = \sqrt{g^2+f^2-c} = \sqrt{(-1)^2+(-1)^2-0} = \sqrt{2}$. The diameter of circle $S$ is $2r = 2\sqrt{2}$.

If a circle $S$ passing through the origin and having its centre on the line $x-y=0$ cuts the circle $x^2+y^2-4x-6y+10=0$ orthogonally,then the diameter of $S$ is

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