$S_1 :$ $x^2+y^2+2 g_1 x+2 f_1 y+c_1=0$

$S_2 :$ $x^2+y^2+2 g_2 x+2 f_2 y+c_2=0$

$S:$ $\ x^2+y^2+2 g x+2 f y+c=0$

$S_1-S_2 =2(g_1-g_2)x +2\left(f_1-f_2\right) y+(c_1-c_2)=0$

$S$ and $S_1$, will also intersect at $P$ and $Q$.

$\therefore$ Redical axis between $S=0$ and $S_1=0$ is $S-S_1=0$

$S-S_1 =2(g-g_1)x +2\left(f-f_1\right) y +c-c_1=0$

$[2(g-g_1) x + 2(f-f_1)y + c - c_1 = 0] \times q$

$2\left(g_1-g_2\right) x+2\left(f_1-f_2\right) y+c_1-c_2=0$

$2 q\left(g-g_1\right) x+2 q\left(f-f_1\right) y+q\left(c-c_1\right)=0$

$\left(g_1-g_2\right)=q\left(g-g_1\right),\left(f_1-f_2\right)=q\left(f-f_1\right)$

and $c_1-c_2=q\left(c-c_1\right)$

$g=\frac{\left(g_1-g_2\right)}{q}+g_1, f=\frac{\left(f_1-f_2\right)}{q}+f_1, \text { and } c=\frac{\left(c_1-c_2\right)}{q}+c_1$

$g=g_1 \frac{(1+q)}{q}-\frac{g_2}{q},$

$f=f_1 \frac{(1+q)}{q}-\frac{f_2}{q} \text { and }$

$c = c_1 \frac{(1+q)}{q}-\frac{c_2}{q} .$

$S: x^2+y^2+2 x\left(g_1 \frac{(1+q)}{q}-\frac{g_2}{q}\right)$

$+2 y\left(f_1 \frac{(1+q)}{q}-\frac{f_2}{q}\right)+\left.c_1\frac{1+q}{q}\right){-\frac{c_2}{q}}=0$

$x^2=\frac{(1+q)}{2}x^2-\frac{1}{q} \cdot x^2$

$S: {\left(1+q\right)}{}\left(x^2+y^2+2 g_1 x+2 f_1 y+c_1\right)$

$-1\left(x^2+y^2+2 g_2 x+2 f_2 y+c_2\right)=0$

$S:(1+q) S_1-S_2=0$

$(1+q)\left(S_1-\frac{1}{1+q} S_2\right)=0$

$(1+q)\left(S_1+\left(\frac{-1}{1+q}\right) S_2\right)=0$

$\quad S_1+k S_2=0, k=\frac{-1}{1+q} \neq-1$

$S_1=x^2+y^2+2 x+4 y-4=0, \quad O_1(-1,-2), \quad \ r_1=3$

$S_2=x^2+y^2+6 y=0,\quad O_2(0,-3), \quad r_2=3$

$\left|r_1-r_2\right| \quad < \quad d_{O_1,O_2}=\sqrt{2} \quad < \quad6$

$S=S_1+k S_2=0 \text { where } k \text { is real and } k\neq{-1}$

Solution:

$S=\left(x^2+y^2+2 x+4 y-4\right)+k\left(x^2+y^2+6 y\right)=0$

$(1+k) x^2+(1+k) y^2+2 x+(4+6k)y -4 =0$

$S^{\prime}=0$

$x^2+y^2+2g^{\prime}x+2f^{\prime}+c^{\prime}=0$

$L=mx+ny+p=0$

Radical axis between $S=0$ and $s^{\prime}=0$ muSt be $L=0,$

$S=x^2+y^2+2 g x+2 f y+c=0$

$S-S^{\prime}=0$, ie;

$S-S^{\prime}=2\left(g-g^{\prime}\right) x+2\left(f-f^{\prime}\right) y+c-c^{\prime}=0$

$L=m x+n y+p=0 \quad \quad \times q$

$m q x + nqy +pq=0$

$m q=2\left(g-g^{\prime})\Rightarrow 2 g=2 g^{\prime}+m q\right.$

$n q=2\left(f-f^{\prime}\right) and / \Rightarrow 2 f=2 f^{\prime}+n q$

$p q=c-c^{\prime} \Rightarrow c=c^{\prime}+p q$

$S=x^2+y^2+2 g x+2 f y+c=0$

$S=x^2+y^2+\left(2 g^{\prime}+m q\right) x+\left(2 f^{\prime}+n q\right) y+ c^{\prime} + pq=0$

$\left(x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}\right)+q(m x+n y +p)=0$

$S=\left( S^{\prime}+q L=0\right)$

$\left[{\because S^{\prime} = \left(x^2+y^2+2 g^{\prime} x+2 f^{\prime} y+c^{\prime}\right), \& \quad L = (m x+n y +p)}\right]$

$S^{\prime}:\left(x-\frac{x_1+x_2}{2}\right)^2 +\left(y-\frac{y_1-y_2}{2}\right)^2 -\frac{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}{4}=0$

$\frac{y-y_1}{x-x_1}=\frac{y_1-y_2}{x_1-x_2}$

$L\left(x-x_1\right)\left(y_1-y_2\right)+\left(y-y_1\right)\left(x_2-x_1\right)=0$

$S=S^{\prime}+K L=0 \quad(K \text { is real })$

$S=\left(x-\frac{x_1+x_2}{2}\right)^2+\left(y-\frac{y_1+y_2}{2}\right)^2-\left[\frac{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}{4}\right] +k\left(\left(x-x_1\right)\left(y_1-y_2\right)+\left(y-y_1\right)\left(x_2-x_1\right)\right)=0$

$(x-4)^2+(y+2)^2=8$

$S^{\prime }=(x-y)^2+(y+2)^2-8=0$

$\text { i.e., } S^{\prime}=x^2+y^2-8 x+4 y+12=0$

$\frac{y-0}{x-2} =\frac{0-(-2)}{2-4} =-1$

$x+y-2=0$

$L=x+y-2=0 .$

$S=S^{\prime}+k L=0$

$x^2+y^2+(k-8) x+(k+4) y+(12-2 k)=0$

$S=x^2+y^2+(k-8) x+ (k+4) y+(12-2 k)=0$

$g^2+f^2-c$

$=\frac{(k-8)^2}{4}+\frac{(k+4)^2}{4}-(12-2 k)$

$=\frac{\left(2 k^2-8 k+80-48+8 k)\right.}{4} =\frac{2 k^2+32}{4}>0$

$4+0 +(k-8) 2 +(12-2 k) = 4+2 k-16+12-2 k$

$=0 .$

Thank you