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?

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    $\begingroup$ Have a look at this image: en.wikipedia.org/wiki/Trigonometric_functions#/media/… $\endgroup$ – Sean Lake Aug 30 '16 at 15:15
  • $\begingroup$ I have seen the image. Thank you for that, but what does it have to do with my question? $\endgroup$ – Carl Terence Valdellon Aug 30 '16 at 15:26
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    $\begingroup$ I was having trouble understanding your question because I was looking at an earlier draft that was not super readable. Adding a diagram, if you can, might increase the likelihood of getting useful answers. $\endgroup$ – Sean Lake Aug 30 '16 at 15:29
  • $\begingroup$ I am sorry, Sean. I would like to add a diagram, if only I knew how. $\endgroup$ – Carl Terence Valdellon Aug 30 '16 at 15:32
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    $\begingroup$ I think this is a mistake, since $B$ is already on the circle: "First, I extended $AB$ to a point on the circle..." (edit: never mind, you mean on the opposite side from $B$). Also, shouldn't $x=BC$? $\endgroup$ – Sean Lake Aug 30 '16 at 15:32

$\qquad\qquad$enter image description here

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.

  • $\begingroup$ should $\cos \angle OAB$ be $\cos \angle OBA$ instead? $\endgroup$ – Mick Aug 31 '16 at 3:11
  • $\begingroup$ @Mick: Yes, thank you. I have corrected it. $\endgroup$ – mathlove Aug 31 '16 at 4:37

By power of the point $C$ respect to the circle of radius $OD$ one has $$\overline{CD}^2=\overline{CB}(\overline{CA}+\overline{AB'})$$ where $B'$ is the other point of intersection of line $CA$ with the circle. Hence $\overline{AB'}=5$.

Let $r$ be the radius of the circle. By Stewart's theorem in $\triangle{BOB'}$ one has $$r^2(4+5)=9(4\cdot5+2^2)\iff r^2=24$$ Now by Pythagoras in $\triangle{CDO}$, $$6^2+r^2=(\overline{OC})^2\iff \color{red}{\overline{OC}=2\sqrt{15}}$$


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