Given a circle with center O:
Segment CD is tangent to the circle with center O, at D. Point A is in the interior of the circle, and segment AC intersects the circle at B. If OA=2, AB=4, BC=3 and CD=6, find the length of segment OC.
Here are the things I tried. First, I extended AB to a point on the circle and solved for the length of the extended segment using tangent-secant formula for circles; I found it to be 5 units. Also, I assigned OC=r+x, where r is the radius of the circle. Next, I concluded that, by Pythagorean Theorem, $$(x+r)^2=r^2+36$$ and thus $$x^2+2rx-36=0$$ must be true. Here, I am stuck. I eventually gave up, and looked at the answer at the back of the questionnaire. The answer is $$2 \sqrt{15}$$ Can anyone help me in solving this problem?
According to the law of cosines, $\cos (\theta )={{r^2+r^2-d^2}\over {2rr}}=1- {{d^2}\over {2r^2}}$. 
Best Answer
$\qquad\qquad$
Let $E\ (\not=B)$ be the point both on the line $AC$ and on the circle. Also, let $F,G$ be the points both on the line $OC$ and on the circle. ($G$ is near to $C$)
As you wrote, $6^2=3\times (3+4+AE)\implies AE=5$. Now since $\triangle{OBE}$ is an isosceles triangle with $OE=OB=r,OA=2,EA=5,BA=4$, applying the law of cosines to $\triangle{OAB}$ and $\triangle{OAE}$ gives
$$\frac{r^2+4^2-2^2}{2\cdot 4\cdot r}=\cos\angle{OBA}=\cos\angle{OEA}=\frac{r^2+5^2-2^2}{2\cdot 5\cdot r}\implies r=2\sqrt 6$$
So, from the equation you have, $$x^2+4\sqrt 6\ x-36\implies x=-2\sqrt 6+2\sqrt{15}$$ from which $$OC=x+r=2\sqrt{15}$$ follows.