<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] Your original equation can be restated as $y^2 = 1 - x^2$.  Using implicit differentiation,
$$  2y y' = -2x x'  \text{.}  $$
Specializing the independent variable to $x$, we obtain
$$  2y \frac{\mathrm{d}y}{\mathrm{d}x} = -2x  \text{,}  $$
so
$$  \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-x}{y}  \text{.}  $$
So let's run our finger, the variable $a$, along the semicircle as $x$ ranges from $0$ to $1$ and see where the tangent line meets the $x$-axis.  We have the point $(a,\sqrt{1-a^2})$ and the slope $\frac{-a}{\sqrt{1-a^2}}$, so the line is
$$  y - \sqrt{1-a^2} = \frac{-a}{\sqrt{1-a^2}}(x-a)  \text{.}  $$
To pass through the point $(2,0)$, we must satisfy
$$  0 - \sqrt{1-a^2} = \frac{-a}{\sqrt{1-a^2}}(2-a)  \text{,}  $$
which we can solve for $a$.  \begin{align*}
    - \sqrt{1-a^2} &= \frac{-a}{\sqrt{1-a^2}}(2-a)  \\
    - (1-a^2) &= -a(2-a)  \\
    a^2 - 1 &= a^2 - 2a  \\
    1 &= 2a  \\
    a &= 1/2  \text{.}
\end{align*}
Therefore, the equation of the line is
$$  y - \sqrt{1-\frac{1}{4}} = \frac{-1/2}{\sqrt{1-\frac{1}{4}}}\left( x-\frac{1}{2} \right)  \text{.}  $$
(This simplifies a little, which I leave to the reader.)
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