If the circle $\mathrm{S}=0$ cuts the circles $x^2+y^2-2 x+6 y=0$. $x^2+y^2-4 x-2 y+6=0$ and $x^2+y^2-12 x+2…
If the circle $\mathrm{S}=0$ cuts the circles $x^2+y^2-2 x+6 y=0$. $x^2+y^2-4 x-2 y+6=0$ and $x^2+y^2-12 x+2 y+3=0$ orthogonally, then equation of the tangent at $(0,3)$ on $S=$ 0 is
$x+y-3=0$
$y=3$
$x=0$
$x-y+3=0$
Solution
$\begin{aligned} Let \quad & S_1 \equiv x^2+y^2-2 x+6 y=0 \\ & S_2 \equiv x^2+y^2-4 x-2 y+6=0\end{aligned}$
and $S_3 \equiv x^2+y^2-12 x+2 y+3=0$
Also let $S \equiv x^2+y^2+2 g x+2 f y+c=0$
Since, $S_1$ and $S$ cut orthogonally
So, $2 g_1 \cdot g_2+2 f_1 \cdot f_2=c_1+c_2$
$\Rightarrow 2 g \cdot(-1)+2 f \cdot(3)=0+c \Rightarrow-2 g+6 f=c...(i)$
Similarly $S_2$ and $S \Rightarrow 2 g(-2)+2 f(-1)=6+c$
$\Rightarrow-4 g-2 f=6+c..(ii)$
and $S_3$ and $S \Rightarrow 2 g(-6)+2 f(1)=3+c$
$\Rightarrow-12 g+2 f=3+c...(iii)$
After solving (i), (ii) and (iii),
we get $g=0, f=\frac{-3}{4}$ and $c=\frac{-9}{2}$
Since, equation of tangent of $S \equiv 0$ is
$\begin{aligned}
& x x_1+y y_1+g\left(x+x_1\right)+f\left(y_1+y\right)+c=0 \\
& \Rightarrow x \times 0+3 \times y+0-\frac{3}{4}(3+y)-\frac{9}{2}=0 \\
& \Rightarrow 3 y-\frac{9}{4}-\frac{3 y}{4}-\frac{9}{2}=0 \Rightarrow y=3
\end{aligned}$