A common tangent to the circle $x^2+y^2=9$ and parabola $y^2=8 x$ is
A common tangent to the circle $x^2+y^2=9$ and parabola $y^2=8 x$ is
$3 x-\sqrt{3} y+2=0$
$x-\sqrt{3} y+6=0$
$2 x-\sqrt{3} y+3=0$
$x-3 y+6=0$
Solution
Let $y=m x+c$ be the equation of common tangent to parabola $y^2=8 x$ and $x^2+y^2=9$
Condition for $y=m x+c$ to be the tangent of $y^2=4 a x$ is
$c=\frac{a}{m}$ So, $c=\frac{2}{m}$ ....(i)
The line $y=m x+c$ is tangent to circle $x^2+y^2=9$
$\Rightarrow \frac{c}{\sqrt{m^2+1}}=3 \Rightarrow \frac{2}{\sqrt{m^2+1}}=3 m$
$\Rightarrow \frac{4}{m^2+1}=9 m^2 \Rightarrow 9 m^4+9 m^2-4=0$
$\Rightarrow m^2=\frac{1}{3}, \frac{-4}{3} \Rightarrow m=\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}} \Rightarrow c=2 \sqrt{3},-2 \sqrt{3}$
$\therefore y=\frac{1}{\sqrt{3}} x+2 \sqrt{3} \Rightarrow x-\sqrt{3} y+6=0$