Let the circle S equiv x^2+y^2+2gx+2fy+c=0 cu | vedclass

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(D) The general equation of the circle is $S \equiv x^2+y^2+2gx+2fy+c=0$ with center $(-g, -f)$.For two circles $x^2+y^2+2g_1x+2f_1y+c_1=0$ and $x^2+y

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$c-f$

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(D) The general equation of the circle is $S \equiv x^2+y^2+2gx+2fy+c=0$ with center $(-g, -f)$.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 be orthogonal,the condition is $2g_1g_2+2f_1f_2=c_1+c_2$.For the first circle $x^2+y^2-2x+2y-2=0$,we have $g_2=-1, f_2=1, c_2=-2$. Applying the condition: $2g(-1)+2f(1)=c-2 \Rightarrow -2g+2f=c-2$ (Equation $1$).For the second circle $x^2+y^2+4x-6y+9=0$,we have $g_3=2, f_3=-3, c_3=9$. Applying the condition: $2g(2)+2f(-3)=c+9 \Rightarrow 4g-6f=c+9$ (Equation $2$).The center $(-g, -f)$ lies on $2x+3y-2=0$,so $2(-g)+3(-f)-2=0 \Rightarrow -2g-3f=2$ (Equation $3$).Solving the system of equations:From $(1)$,$c = -2g+2f+2$.Substitute into $(2)$: $4g-6f = (-2g+2f+2)+9 \Rightarrow 6g-8f = 11$.From $(3)$,$2g = -3f-2$.Substitute $2g$ into $6g-8f=11$: $3(-3f-2)-8f=11$ $\Rightarrow -9f-6-8f=11$ $\Rightarrow -17f=17$ $\Rightarrow f=-1$.Then $2g = -3(-1)-2 = 1 \Rightarrow g=1/2$.Then $c = -2(1/2)+2(-1)+2 = -1-2+2 = -1$.We need $2g+f = 2(1/2)+(-1) = 1-1 = 0$.Checking the options: $c-f = -1-(-1) = 0$. Thus,$2g+f = c-f$.

Let the circle $S \equiv x^2+y^2+2gx+2fy+c=0$ cut the circles $x^2+y^2-2x+2y-2=0$ and $x^2+y^2+4x-6y+9=0$ orthogonally. If the centre of the circle $S=0$ lies on the line $2x+3y-2=0$,then $2g+f=$

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