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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

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  • $\begingroup$ Have you made an image? $\endgroup$ – Dr. Sonnhard Graubner Aug 29 '19 at 11:15
  • $\begingroup$ No, very sorry. $\endgroup$ – user690234 Aug 29 '19 at 11:20
  • $\begingroup$ No, because the two lines are not collinear $\endgroup$ – user690234 Aug 29 '19 at 11:27
  • $\begingroup$ Why is there someone called Anonymous Leopard on my drawing? $\endgroup$ – user690234 Aug 29 '19 at 11:35
  • $\begingroup$ That is just a generic name: all the email addresses are anonymised to protect privacy. $\endgroup$ – Toby Mak Aug 29 '19 at 11:48
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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: $$r^2 = 4-24\cos(\theta) = (2)(13-12\cos(\theta)) - 22= 2r^2 - 22$$ $$r = \sqrt{22}$$

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

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

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}

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

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