<s>[INST] 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.
 [/INST] 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 $(x,y)$
$\frac {y-0}{x-2} = -\frac {x}{y}\\
y^2 = -x^2 + 2x\\
1-x^2 = -x^2 + 2x\\
x = \frac {1}{2}$
$y = -\frac {1}{\sqrt 3}(x-2)$
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