Segment AB is tangent at A to the circle with center O,point D is interior to the circle, and segment DB intersects the circle at C. If BC=CD=3, OD=2, and AB=6, then find the radius of the circle.
Currently, I have thought about ways to use trigonometry to solve this problem, but none of them worked. Can you please help me tackle this problem?
https://docs.google.com/drawings/d/11ll_gefQekm1MgBicdkBzvS17b8UOSp8MlFU9WN5Z4A/edit?usp=sharing
On
$OC$ is a median of $\Delta ODB.$
Thus, $$OC=\frac{1}{2}\sqrt{2OD^2+2OB^2-DB^2}$$ or $$r=\frac{1}{2}\sqrt{2\cdot2^2+2(r^2+36)-6^2}.$$ Can you end it now?
I got $r=\sqrt{22}.$
I used the following.
In parallelogram $ABCD$ we know that: $$AB^2+BC^2+CD^2+DA^2=AC^2+BD^2.$$ Let $AB=c$, $BC=a$, $AC=b$ and $BD=2m_b$, where $m_b$ be a median to $AC$ of $\Delta ABC$ .
Thus, $$2(a^2+c^2)=b^2+4m_b^2,$$ which gives $$m_b=\frac{1}{2}\sqrt{2a^2+2c^2-b^2}.$$
On
Extend $BD$ to get the point $E$ at another intersection with the circle.
Using the power of the point $B$ with respect to the circle, we have
\begin{align} |AB|^2& =|BE|\cdot|BC| =(|BD|+|DE|)\cdot|BC| ,\\ 6^2&=(6+|DE|)\cdot 3 ,\\ |DE|&=6 . \end{align}
By the Stewart’s Theorem wrt $\triangle OCE$,
\begin{align} |OE|^2\cdot|CD|+|OC|^2\cdot|DE| &= (|CD|+|DE|)\cdot(|OD|^2+|DE|\cdot|CD|) ,\\ R^2(3+6)&= (3+6)\cdot(2^2+3\cdot 6) ,\\ R^2&=22 . \end{align}
Let $\theta = ∠ODB$. We have 2 ways to get radius:
Law of Cosine on ΔODC: $$r^2 = 2^2 + 3^2 - 2(2)(3)\cos(\theta) = 13-12\cos(\theta)$$
Law of Cosine on ΔODB: $$OB^2 = 2^2 + 6^2 - 2(2)(6)\cos(\theta) = 40-24\cos(\theta)$$ $$r^2 = OB^2 - AB^2 = 4-24\cos(\theta)$$ Combine both ways: $$r^2 = 4-24\cos(\theta) = (2)(13-12\cos(\theta)) - 22= 2r^2 - 22$$ $$r = \sqrt{22}$$