Question

Equation of a common tangent to the circle, $x^2+y^2-6x=0$ and the parabola, $y^2=4x$, is

• A. $2\sqrt{3}y=-x-12$
• B. $2\sqrt{3}y=12x+1$
• C. $\sqrt{3}y=3x+1$
• D. $\sqrt{3}y=x+3$

Answer 1

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

The curves are $y^2=4x$
and $x^2+y^2-6x=0\Rightarrow (x-3{)}^2+y^2=9$
Equation of tangent to $y^2=4x$ is
$y=mx+\frac{1}{m}...\left(1\right)$
Equation of tangent to the circle is
$y=m(x-3)+3\sqrt{1+m^2}$,
where $m$ is the slope.
Since the two curves have common tangents,
$\Rightarrow mx+\frac{1}{m}=m(x-3)+3\sqrt{1+m^2}$
$\Rightarrow m=\pm \frac{1}{\sqrt{3}}$

Putting the value of $m$ in $\left(1\right)$, we get
$\sqrt{3}y=\pm (x+3)$
i.e., $\sqrt{3}y=x+3$