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.

If we construct the radius from the center of the circle to the point of intersection, this radius, the line and the x axis forms a right triangle with angle $\theta$ at the origin.
$\sec \theta = 2$
The point of intersection is $(\cos \theta, \sin\theta) = (\frac 12, \frac {\sqrt 3}{2})$
The line intersects the y-axis at $(0,\csc \theta) = \frac {2}{\sqrt 3}$
In intercept-intercept form: $\frac {x}{2} + \frac {y\sqrt 3}{2} = 1$
Or in standard form  $x + y\sqrt 3 = 2$
If you really want to use calculus, we have:
$\frac {dy}{dx} = -\frac {x}{\sqrt {1-x^2}} = -\frac {x}{y}$
And this slope must equal the slope of the line from $(2,0)$ to $(x,y)$
$\frac {y-0}{x-2} = -\frac {x}{y}\\
y^2 = -x^2 + 2x\\
1-x^2 = -x^2 + 2x\\
x = \frac {1}{2}$
$y = -\frac {1}{\sqrt 3}(x-2)$
