# Circle problem: What is the radius

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?

• Have you made an image? – Dr. Sonnhard Graubner Aug 29 '19 at 11:15
• No, very sorry. – user690234 Aug 29 '19 at 11:20
• No, because the two lines are not collinear – user690234 Aug 29 '19 at 11:27
• Why is there someone called Anonymous Leopard on my drawing? – user690234 Aug 29 '19 at 11:35
• That is just a generic name: all the email addresses are anonymised to protect privacy. – Toby Mak Aug 29 '19 at 11:48

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: $$r^2 = 4-24\cos(\theta) = (2)(13-12\cos(\theta)) - 22= 2r^2 - 22$$ $$r = \sqrt{22}$$

$$OC$$ 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 $$m_b=\frac{1}{2}\sqrt{2a^2+2c^2-b^2}.$$

• How did you get the first equation about OC? – user690234 Aug 29 '19 at 11:33
• By the way even if the first equation is correct, r is about 1.981, which is impossible. – user690234 Aug 29 '19 at 11:41
• @user690234 I added something. See now. – Michael Rozenberg Aug 29 '19 at 11:42
• @user690234 I fixed the typo. See again. – Michael Rozenberg Aug 29 '19 at 11:48
• I can't see any difference with before. – user690234 Aug 29 '19 at 11:52 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}

• what is stewart's theorem? – user690234 Aug 30 '19 at 9:49
• @user690234: Stewart's theorem is a nice useful tool to have along with the sine/cosine rules. – g.kov Aug 30 '19 at 12:10