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 $\sqrt{3}y=x+3$

Any tangent to $y^2=4x$ is of the form $y=mx+\frac{1}{m}$, $(\because a=1)$ and this touches the circle $(x-3{)}^2+y^2=9$.
If $\left|\frac{m\left(3\right)+\frac{1}{m}-0}{\sqrt{m^2+1}}\right|=3$
[$\because$ centre of the circle is (3, 0) and radius is 3].
$\Rightarrow \frac{3m^2+1}{m}=\pm 3\sqrt{m^2+1}$
$\Rightarrow 3m^2+1=\pm 3m\sqrt{m^2+1}$
$\Rightarrow 9m^4+1+6m^2=9m^2(m^2+1)$
$\Rightarrow 3m^2=1$
$\Rightarrow m=\pm \frac{1}{\sqrt{3}}$
If the tangent touches the parabola and circle above the X-axis, then slope m should be positive.
$\therefore m=\frac{1}{\sqrt{3}}$ and the equation is $y=\frac{1}{\sqrt{3}}x+\sqrt{3}$
or $\sqrt{3}y=x+3$.