Help with finding the equation of line tangent to the semi circle with the equation: $y=\sqrt{1-x^2}$ 
Ok so I need to show that the equation of the tangent above to the semi-circle with the equation: $y=\sqrt{1-x^2}$,  is $y=-\frac{1}{\sqrt{3}}x$ + $\frac{2}{\sqrt{3}}$
What we from the question:
The tangent intersects the $x$-axis at $(2,0)$, (assume that you do not know any other points of intersection)
What I tried to do so far:
Since its a tangent, I decided to differentiate the semi-circle function using the chain rule, getting:
$\frac{dy}{dx}$ = $\frac{-x}{\sqrt{1-x^2}}$
Then using the equation of a line formula:
$y=mx+c$
--> $y=\frac{-x}{\sqrt{1-x^2}}x +c$
Substituting the point in:
-->$0=\frac{-2}{\sqrt{1-2^2}}(2) +c$
Now clearly something is wrong because I will get the square root of a negative, which does not make any sense. Need help from here.
*Note: I know that there are other ways to solve this, but I would prefer if calculus was used to solve this question.


$OA = (\cos t, \sin t)$ and $AB = (2 - \cos t, -\sin t)$. For $OA \perp AB$, we need the dot product $\cos t(2 - \cos t) + \sin t \cdot - \sin t = 0 \implies 2 \cos t - 1 = 0, t = \frac{\pi}{3}$.
Thus the line segment $AB = \left(\frac{3}{2}, -\frac{\sqrt3}{2} \right)$, which has slope $-\frac{\sqrt3}{3}$ and passes through $(\cos \pi/3, \sin \pi/3) = (1/2, \sqrt3/2)$.

Here is a non-calculus method for reference:
We know that $OA = 1, OB = 2$ and $\angle OAB$ is right since it is tangent to the circle.
Thus $\cos \angle AOB = \frac{1}{2} \implies \angle AOB = \pi/3$. Thus the gradient of $OA$ is $\text{rise}/\text{run} = \tan \pi/3 = \sqrt{3}$, and so $AB$ has a gradient of $-\frac{1}{\sqrt{3}}$.
$AB$ must pass through $A$ which is $(\cos \pi/3, \sin \pi/3) = (1/2, \sqrt3/2)$, therefore the equation is:
$$y - \sqrt3/2 = -\frac{1}{\sqrt3} \left(x - \frac{1}{2} \right) \implies y = -\frac{1}{\sqrt3}x + \frac{1 + (\sqrt{3})^2}{2\sqrt3}$$
