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

Let equation of tangent to the parabola $y^2=4x$ is

$y=mx+\frac{1}{m}$

$m^{2}x-ym+1=0$ is tangent to $x^2+y^2-6x=0$

$⟹\frac{|3m^2+1|}{\sqrt{m^4+m^2}}=3$

$m=\pm \frac{1}{\sqrt{3}}$

Therefore,

Tangents are $x+\sqrt{3}y+3=0$ and $x-\sqrt{3}y+3=0$.

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