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.
 A: 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 $(x,y)$
$\frac {y-0}{x-2} = -\frac {x}{y}\\
y^2 = -x^2 + 2x\\
1-x^2 = -x^2 + 2x\\
x = \frac {1}{2}$
$y = -\frac {1}{\sqrt 3}(x-2)$
A: 
$OA = (\cos t, \sin t)$ 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:
$$y - \sqrt3/2 = -\frac{1}{\sqrt3} \left(x - \frac{1}{2} \right) \implies y = -\frac{1}{\sqrt3}x + \frac{1 + (\sqrt{3})^2}{2\sqrt3}$$
A: You correctly found that the derivative is
$$y' = -\frac{x}{\sqrt{1 - x^2}}$$
The point-slope form of the equation of the tangent line is
$$y - y_0 = m(x - x_0)$$
where $m$ is the slope and $(x_0, y_0)$ is a point on the tangent line.  Since the tangent line passes through the point $(2, 0)$, it satisfies the equation
$$y - 0 = m(x - 2)$$
Substituting $-\dfrac{x}{\sqrt{1 - x^2}}$ for $m$ and $\sqrt{1 - x^2}$ for $y$ yields
$$\sqrt{1 - x^2} = -\frac{x}{\sqrt{1 - x^2}}(x - 2)$$
Solve the above equation for $x$ in order to find the $x$-coordinate of the point of tangency, then substitute that value of $x$ into the equation $y = \sqrt{1 - x^2}$ to find its $y$-coordinate.  The slope of the tangent line is found by evaluating the derivative at the point of tangency.
A: 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.)
