Question

The equation of the common tangent touching the circle $(x-3{)}^2+y^2=9$ and the parabola $y^2=4x$ above the $x-$axis is

• A. $\sqrt{3}y=3x+1$
• B. $\sqrt{3}y=-(x+3)$
• C. $\sqrt{3}y=x+3$
• D. $\sqrt{3}y=-(3x+1)$

Answer 1

The correct option is C $\sqrt{3}y=x+3$

Equation of a tangent to the parabola $y^2=4x$ will be $y=mx+\frac{1}{m}$
$my=m^{2}x+1$

$my-m^{2}x-1=0$
The line is also tangent to the circle. Hence, its distance from the centre

$(3,0)$ will be equal to the radius $3$.

Hence,
$\frac{3m^2+1}{\sqrt{m^4+m^2}}=3$
On squaring,

$9m^4+6m^2+1=9m^4+9m^2$
$\Rightarrow 3m^2=1$
$m=\pm \frac{1}{\sqrt{3}}$

Hence the equations of the common tangent are :
$y=\frac{x}{\sqrt{3}}+\sqrt{3}$
$\sqrt{3}y=x+3$

The other tangent is
$y=\frac{-x}{\sqrt{3}}-\sqrt{3}$
$\sqrt{3}y+x+3=0$. This common tangent is not above the X axis.
Hence, option C is correct