Suppose the circle S: x^2+y^2+2gx+2fy+c=0 cut | vedclass

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(C) The condition for two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ to cut orthogonally is $2g_1g_2+2f_1f_2=c_1+c_2$. For $S

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$4$

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(C) The condition for two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y^2+2g_2x+2f_2y+c_2=0$ to cut orthogonally is $2g_1g_2+2f_1f_2=c_1+c_2$. For $S$ and $S'$,we have $2g(-2)+2f(-3)=c+11 \implies -4g-6f=c+11$ $(i)$. For $S$ and $S''$,we have $2g(-5)+2f(-2)=c+21 \implies -10g-4f=c+21$ $(ii)$. The centre of $S$ is $(-g, -f)$. Since it lies on the bisector of the angle between positive coordinate axes $(y=x)$,we have $-f = -g$,so $f=g$ $(iii)$. Substituting $f=g$ into $(i)$ and $(ii)$: $-10f = c+11$ $(iv)$ $-14f = c+21$ $(v)$ Subtracting $(v)$ from $(iv)$: $4f = -10 \implies f = -2.5$. Thus $g = -2.5$. From $(iv)$,$c = -10(-2.5) - 11 = 25 - 11 = 14$. Finally,$2g+2f+c = 2(-2.5)+2(-2.5)+14 = -5-5+14 = 4$.

List-$I$	List-$II$
$(A)$ $(x-2)^2+y^2=2$,$(x-2)^2+(y-1)^2=1$	$I.$ $90^{\circ}$
$(B)$ $x^2+y^2-6x-6y+9=0$,$x^2+y^2-4x+4y-9=0$	$II.$ $135^{\circ}$
$(C)$ $x^2+y^2+4x-14y+28=0$,$x^2+y^2+4x-5=0$	$III.$ $60^{\circ}$
	$IV.$ $30^{\circ}$

Suppose the circle $S: x^2+y^2+2gx+2fy+c=0$ cuts orthogonally the two circles $S': x^2+y^2-4x-6y+11=0$ and $S'': x^2+y^2-10x-4y+21=0$. If the centre of $S=0$ lies on the bisector of the angle between the positive coordinate axes,then $2g+2f+c=$

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