The straight line touching the circle $x^2+y^2-2 x-3=0$ and remaining normal to the circle $x^2+y^2-4 y-6=0$…
The straight line touching the circle $x^2+y^2-2 x-3=0$ and remaining normal to the circle $x^2+y^2-4 y-6=0$ is
$4 x-3 y+6=0$
$y+2=0$
$4 x+3 y-6=0$
$2 x+3=0$
Solution
Line is normal to circle $x^2+y^2-4 y-6=0$
$\therefore$ It passes through centre $(0,2)$.
Let $m$ be slope of line.
$\therefore$ Equation of line is $y-2=m x$
$
\Rightarrow \quad m x-y+2=0
$
This line is tangent to $x^2+y^2-2 x-3=0$
centre is $(1,0)$, radius is 2
$
\begin{aligned}
& \therefore \quad 2=\left|\frac{m+2}{\sqrt{1+m^2}}\right| \\
& \Rightarrow \quad 4\left(1+m^2\right)=(m+2)^2 \\
& \therefore \quad 3 m^2-4 m=0 \\
& \Rightarrow \quad m=0 \\
& \text { or } \quad m=\frac{4}{3} \\
&
\end{aligned}
$
$\therefore \quad$ Line can be $y=2$
or $\quad 4 x-3 y+6=0$