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

Question

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

Accepted Answer

Circle: centre $(3,0)$, $r=3$. Tangent to parabola: $y=mx+\dfrac1m$. Distance from $(3,0)$ equals $3$:
$$\frac{|3m+1/m|}{\sqrt{m^2+1}}=3\Rightarrow9m^2+6+\frac1{m^2}=9m^2+9\Rightarrow m^2=\frac13.$$

$$y=\frac x{\sqrt3}+\sqrt3\ \Longrightarrow\ \sqrt3y=x+3.$$

Solutions · 1