Let x+y=0 be the radical axis of the circles | vedclass

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(B) The radical axis of two circles $S=0$ and $S'=0$ is given by $S-S'=0$. Given $S \equiv x^2+y^2+2gx+2fy+c=0$ and $S' \equiv x^2+y^2-6x-4y+4=0$. Sub

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$\pm 3$

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(B) The radical axis of two circles $S=0$ and $S'=0$ is given by $S-S'=0$. Given $S \equiv x^2+y^2+2gx+2fy+c=0$ and $S' \equiv x^2+y^2-6x-4y+4=0$. Subtracting the two equations: $(2g+6)x + (2f+4)y + (c-4) = 0$. Comparing this with the given radical axis $x+y=0$,we have $\frac{2g+6}{1} = \frac{2f+4}{1} = \frac{c-4}{0}$. From $\frac{c-4}{0}$,we get $c-4=0$,so $c=4$. From $\frac{2g+6}{1} = \frac{2f+4}{1}$,we get $2g+6 = 2f+4$,which simplifies to $2g-2f = -2$,or $g-f = -1$. Also,the radius of circle $S$ is $1$,so $\sqrt{g^2+f^2-c} = 1$,which means $g^2+f^2-4 = 1$,or $g^2+f^2 = 5$. Since $f = g+1$,substitute into the radius equation: $g^2 + (g+1)^2 = 5$. $g^2 + g^2 + 2g + 1 = 5 \implies 2g^2 + 2g - 4 = 0 \implies g^2 + g - 2 = 0$. $(g+2)(g-1) = 0$,so $g=1$ or $g=-2$. If $g=1$,$f=2$,so $g+f=3$. If $g=-2$,$f=-1$,so $g+f=-3$. Thus,$g+f = \pm 3$.

List-$I$	List-$II$
$(A)$ Point circles of $S_\alpha=0$	$(i)$ do not exist
$(B)$ Point circles of $S_\beta=0$	(ii) intersecting
$(C)$ The circles in $S_\alpha=0$ are	(iii) non-intersecting
$(D)$ The circles in $S_\beta=0$ are	(iv) $(\pm \sqrt{k}, 0)$
	$(v)$ $(0, \pm \sqrt{k})$

Let $x+y=0$ be the radical axis of the circles $S \equiv x^2+y^2+2gx+2fy+c=0$ and $S' \equiv x^2+y^2-6x-4y+4=0$. If the radius of the circle $S=0$ is $1$,then find the value of $g+f$.

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