The circle $\mathrm{S} \equiv x^2+y^2-2 x-4 y+1=0$ cuts the $y$-axis at $\mathrm{A},…
The circle $\mathrm{S} \equiv x^2+y^2-2 x-4 y+1=0$ cuts the $y$-axis at $\mathrm{A}, \mathrm{B}(\mathrm{OA}\gt\mathrm{OB})$. If the radical axis of $\mathrm{S} \equiv \mathrm{O}$ and $\mathrm{S}^{\prime} \equiv x^2$ $+y^2-4 x-2 y+4=0$ cuts the $y$-axis at $C$ then the ratio in which C divides AB is
$7+2 \sqrt{3}:-7+2 \sqrt{3}$
$\sqrt{3}+2: \sqrt{3}-2$
$6-2 \sqrt{3}: 2 \sqrt{3}-6$
$-3: \sqrt{3}$
Solution
Given, $S \equiv x^2+y^2-2 x-4 y+1=0$
If $x=0 \Rightarrow y^2-4 y+1=0$
$\Rightarrow y=\frac{4 \pm \sqrt{16-4}}{2}=2 \pm \sqrt{3}$
So, $A(0,2+\sqrt{3})$ and $B(0,2-\sqrt{3})$
Now, equation of radical axis is $S-S^{\prime}=0$
$\Rightarrow 2 x-2 y-3=0$
If $x=0 \Rightarrow-2 y=3 \Rightarrow y=\frac{-3}{2}$ So, $C\left(0, \frac{-3}{2}\right)$
Let the required ratio be $K: 1$
$\begin{aligned}
& \text { So, } \frac{-3}{2}=\frac{K(2-\sqrt{3})+1(2+\sqrt{3})}{K+1} \\
& \Rightarrow-3 K-3=4 K-2 K \sqrt{3}+4+2 \sqrt{3} \\
& \Rightarrow K(-7+2 \sqrt{3})=7+2 \sqrt{3} \\
& \Rightarrow K: 1=(7+2 \sqrt{3}):(-7+2 \sqrt{3})
\end{aligned}$