Using Power-of-a-Point Theorem in Geometric Proofs

[theJorge551]

Homework Statement

Point A is on a circle whose center is O, AB is a tangent to the circle, AB = 6, D is inside of the circle, OD = 2, DB intersects the circle at C, and BC = DC = 3. Find the radius of the circle.

Homework Equations

Power of a point theorem (several cases found online, a few here: http://www.cut-the-knot.org/pythagoras/PPower.shtml )

The Attempt at a Solution

I've drawn the figure, and I recognize that in drawing D, it'd have to lay on the edge of a(n) (imaginary) circle about B with a radius of 6, and that based on that, D could only have 2 possible positions inside the original circle. I've also found that the power of point B w.r.t. the original circle is 36, but I don't know how to proceed to find r (either AO or CO, as far as I can tell).

 

[bacon]

TheJorge551, this is a cool problem. I had not seen the Power of a Point theorem before, it’s amazing.
I might be able to help.
So far what you’ve done looks right to me. What I think you should do next is continue the line BD until it intersects the the circle at the far side. I called that point of intersection E. Now you have the line BE that passes through points C and D. You can find the length of the segment DE(with help from Power of the Point theorem). Once you have DE, consider how you might use the Power of a Point theorem yet again.
Hope this helps.

 

[theJorge551]

Thanks for your hint! I did what you suggested, and stated that 36 = BC x BE, And knowing that BC = 2, BE must be 12. Because BD = 6, BE = 6 as well...then I said that there is a diameter through OD intersecting the circle at points F and G (which are both irrelevant, directly, because I can just use "r"), and from the P.o.P. theorem, CD x DE = (r-DO) x (r). Because DO = 2, I solved the quadratic and found r to equal 1 + sqrt(19). Is this the solution you found?

 

[bacon]

We almost agree at this point.
First, you might have a typo here: "And knowing that BC = 2, BE must be 12. Because BD = 6, BE = 6 as well...". For the second BE, I think you mean DE, is that right?

And at the last, "CD x DE = (r-DO) x (r)", I think instead of r as the last term you should have (r+DO). The focal point to apply the Power of the Point theorem is D and so the line segment FG is broken at D into a segment of length r+DO and the other of r-DO. Would you agree?

 

[theJorge551]

You're correct for the first part; it was a typo. For the second part, you're right -- I made a mistake as to what my point of focus was, so it should be CD x DE = (r-DO) x (r+DO), simplifying to 18 = r^2 - 4, and thus r = sqrt(22). Did I make any other silly errors, or does this match what you have?

 

[bacon]

That's the answer I got.

 

[theJorge551]

Fantastic, thanks for your help, bacon. :D

I've got another problem, which I think I've solved, but the proof seems so utterly simple that I'm skeptical of whether or not I've overlooked some big assumption that I might have made. Could someone tell me if my proof is sufficient (if we take the Power-of-a-point theorem, from earlier, to be true)?

"Two circles intersect at distinct points P and Q, and K is on PQ. A line through K forms chord AKB in one circle, and chord CKD in the other. Prove that AK x KB = CK x KD."

I simply stated that by the P.o.P. theorem, PK x KQ = AK x KB, and PK x KQ = CK x KD, and hence AK x KB = CK x KD. Is my "proof" oversimplified? It's certainly not rigorous, but I'm certainly not going to reprove the fundamental theorem over and over again, in each one in the problem set. :P

I've got another problem that I'm trying to plow through, but I'll post it later if I can't see the solution.