Don't find the intersection

Implicitly differentiating the parabola's equation:

$2y\frac{dy}{dx}=4 \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{x}}$

Then taking a point $(a, 2\sqrt{a})$ on the parabola, a tangent is formed:

$y - 2\sqrt{a} = \frac{1}{\sqrt{a}}(x-a)$

$y = \sqrt{a} + \frac{x}{\sqrt{a}}$

Intersecting this with the circle by substitution:

$(x-3)^{2}+(\sqrt{a} + \frac{x}{\sqrt{a}})^2 = 9$

$(1+\frac{1}{a})x^2-4x+a=0$

For the intersection to be tangential, discriminant is zero. Hence:

$16-4(a)(1+\frac{1}{a})= 0$

$a = 3$

Substituting $a=3$ back into the tangent equation:

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

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

Let $m$ be the slope of the tangent. Equation of tangent to the given parabola with slope as $m$ is $y=mx+\frac{1}{m}$ and that to the given circle is $y=m(x-3)+3\sqrt{1+{m}^2}$ . Equating the constant term in both the equations, we get $m=\frac{1}{\sqrt{3}}$ . Substituting the value in any of the above two equations, we get $\huge\color{#3D99F6}{\boxed{\sqrt{3}y=x+3}}$

Refer conic sections in JEE to know about the equation of tangents.

1.- La recta es positiva. 2.- El punto donde corta al eje Y es mayor a 1 y menor a 2. 3.- solo una cumple lo anterior