Find the length of $BC$. 
For a circle centered at $O$, line $L$ is tangent to the circle at $A$ and has slope $-1/2$. If $AB$ is perpendicular to $OC$ and $OB$ has length $2$, then what is the length of $BC$?
 A: Let corner angle at $ C = \alpha$
$$ \tan \alpha = \frac12;\, BC = x ;\, AC= x \sec \alpha $$
By a standard property of right angled triangles,
$$ (x \sec \alpha)^2 = x (x +2) $$
simplify
$$ x = BC = \frac {2}{{\tan^2 \alpha}} = 8.  $$
The sketch is not proportionately drawn.Also
$$ AB =4, AC = 4 \sqrt 5 $$
A: This is a sneaky problem! Here's my illustration:

We were told $\overline{A B}$ is perpendicular to $\overline{O C}$. Therefore, $\angle OBA = \angle ABC = 90°$.
Because line $L$ is tangent to the circle at $A$, and $O$ is the center of the circle, $\angle OAC$ is $90°$.
Because triangles $\triangle OAC$ and $\triangle OAB$ are similar (being right triangles and sharing two vertices and so on), we know that $\varphi = \angle OCA = \angle BAO$, and that
 $$\frac{OA}{OB} = \frac{OC}{OA} \implies OC = OA \frac{OA}{OB}$$
and therefore
$$BC = OC - OB = OA \frac{OA}{OB} - OB = \frac{(OA)^2 - (OB)^2}{OB}$$
Since
$$\sin \varphi = \frac{OB}{OA} \implies OA = \frac{OB}{\sin \varphi}$$
we know that
$$BC = \frac{\left(\frac{OB}{\sin\varphi}\right)^2 - \left(OB\right)^2}{OB} = \frac{OB \; (\cos\varphi)^2}{(\sin\varphi)^2} = \frac{OB}{(\tan\varphi)^2}$$
Here we come to the sneaky part. The slope of $L$ does not matter, because we were not given any information on the orientation of the other elements!
If we assume that "the slope of $L$ is $-1/2$" is meant as with respect to $\overline{OC}$ -- or, equivalently, that $\overline{OC}$ is parallel to the $x$-axis --, then $\tan \varphi = \lvert -1/2 \rvert = 1/2$, and
$$BC = \frac{OB}{(\tan\varphi)^2} = \frac{2}{1/4} = 8$$
which matches what Nikunj commented and Narasimhan answered before me.
Personally, I'm uncomfortable with making such an assumption (based on an unreliable figure, and just to provide an answer), but then again, I'm an ornery fellow anyway.
A: From Euclid we know that $2\cdot BC=(AB)^2$, on the other hand we know that $2AB=BC$, hence $ BC=8$.
A: Let $\theta$ be the angle between $L$ and OC. We have the following equations:
$$\tan(\theta) = \frac{|OA|}{|AC|}=\frac{2}{|AB|},$$
and
$$\frac{|AB|}{|BC|}=\frac{1}{2}=\tan(\theta).$$
Since $\tan(\pi-\theta) = -\tan(\theta)$, we get that $|AB| = 4$ and $|BC| = 8$.
