A curve C passes through origin and the normal at any point passes through (1,0). How to find the common tangent of this curve and parabola $y^{2}=4x$

Question

I figured out the equation of normal by using equation of line and it is

$$y=-\frac{dx}{dy}(x-1)$$

Using this I found the equation of curve which is

$$`y^{2}+x^{2}-2x=0$$

Then for common tangent I equated the $$\frac{dy}{dx}$$ of parabola and curve and I'm getting $x=1/3$

But the answer I have says tangent is $x=0$ What am I doing wrong?

Answer 1

I am not sure how you got $x = 1/3$. From the graphs of the curves, it is clear that the common tangent is $x=0$