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

• theJorge551
In summary: In the meantime, here's a problem set I think might be easier for you to solve.In summary, Jorge551 is trying to solve a problem that he had not seen before, and he is trying to use the Power of a Point theorem.
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).

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.

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?

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?

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?

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.

What is the Power-of-a-Point Theorem?

The Power-of-a-Point Theorem is a geometric theorem that states that if a line is drawn from a point outside a circle and intersects the circle at two points, the product of the lengths of the segments of the line is equal to the square of the length of the line that connects the points of intersection.

How do you use the Power-of-a-Point Theorem in geometric proofs?

The Power-of-a-Point Theorem is used in geometric proofs to help establish relationships between different line segments and angles in a circle. By using the theorem, you can solve for unknown values and make connections between different parts of a circle.

What are the main applications of the Power-of-a-Point Theorem?

The Power-of-a-Point Theorem has several applications in geometry, including determining the location of the center of a circle, proving theorems related to tangents and secants, and solving for unknown values in geometric problems.

Are there any limitations to using the Power-of-a-Point Theorem?

While the Power-of-a-Point Theorem is a useful tool in geometric proofs, it has some limitations. It can only be used for circles and lines that intersect at two points, and it does not apply to other types of shapes or curves.

Why is the Power-of-a-Point Theorem important for scientists?

The Power-of-a-Point Theorem is important for scientists because it allows for the precise calculation of values in geometric problems, which is essential in fields such as engineering, physics, and astronomy. It also helps establish relationships between different parts of a circle, which can be useful in understanding and analyzing various phenomena in the natural world.

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