Investigation: Chord between two Concentric Circles

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Suppose that we have two concentric (sharing the same center) circles, one bigger than the other. We draw a chord of the bigger circle such that it is tangent to the smaller circle. If the length of the chord is 88, then what is the area in between the two circles? (Fun fact: the area between two concentric circles is called an annulus)

You may be asking me now: Isn't there supposed to be more information? You didn't give me the radii of any of the circles, how are you expecting me to solve this?

Well, let's see what we can solve; maybe that will put forth some insight on what we still need.

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First, denote the center to be OO. Let the point of tangency of the chord on the little circle be AA, and an endpoint of the chord be BB. Finally, let the radius of the small circle be rr and the radius of the large circle be RR.

Notice that AB=4AB=4, because ABAB is half of the chord, which has length 88. Since OAB=90\angle OAB=90^{\circ} because ABAB is tangent to the small circle, we can use the Pythagorean Theorem on ΔOAB\Delta OAB.

Using it gives r2+42=R2r^2+4^2=R^2, or 16=R2r216=R^2-r^2. But now it seems like we're stuck. Not only do we not know rr, but we also do not know RR. What should we do?

You Algebra-savvy people out there might go aha! we have difference of squares, so why not factorize? But always remember to check what your final goal is before irrationally doing random computations. We want to find the area in between the two circles. The area of the large circle is πR2\pi R^2, and the area of the small circle is πr2\pi r^2; therefore the area in between is πR2πr2\pi R^2-\pi r^2.

But wait a minute! πR2πr2=π(R2r2)\pi R^2-\pi r^2=\pi (R^2-r^2), and we know the value of R2r2=16R^2-r^2=16! So therefore the area in between the two circles is π(16)=16π\pi (16)=\boxed{16\pi} and we are done.

Now, this is pretty neat! Without knowing the radii of either circle--just knowing the length of the chord--let us solve the area between the circles. Does that mean that the radius of the circles can be anything and the area in between won't change?

Well, why don't you try it yourself? Change rr into anything you want: 33, 20142014, even 2+π\sqrt{2}+\pi; and see if the area in between changes or not.

A special case of rr might have come up to your mind now: what if r=0r=0?

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In that case, the circle in the center disappears, and the problem degenerates to finding the area of a circle with diameter 88! And when we find the area, we see that indeed, the area still equals 16π16\pi. This strategy--of considering a degenerate, or special, case of a more general problem--is often useful for solving questions faster. Finding the area of a simple circle certainly is easier than doing all that we just did; however, use this strategy with caution.

You've been listening to me rambling on for quite some time now; why not try a few questions for yourself?

PROBLEMS

1. Two concentric circles have the property that a chord of the bigger circle that is tangent to the smaller one has length 1010. What is the area in between the two circles?

2. Three concentric circles A,B,CA,B,C have the property that the radius of AA is bigger than the radius of BB, and the radius of BB is bigger than the radius of CC. A chord of circle BB of length 88 is drawn such that it is tangent to circle CC; a chord of circle AA of length xx is drawn such that it is tangent to circle BB. If the area between circle AA and BB is twice the area between circle BB and circle CC, then what is the value of xx? Express your answer in simplest radical form.

3. A regular pentagon is both circumscribed by a circle and inscribed by a circle. Let the area between the circles be AA. A regular decagon is also both inscribed and circumscribed by a circle. Let the area between those circles be BB. If the side lengths of the two polygons are equal, then what is the ratio of A:BA:B?

4. Prove the general formula. That is, the area between two circles, with the chord of one tangent to another, is πx24\dfrac{\pi x^2}{4}, where xx is the length of the chord.

Feel free to ask any questions, or post your own problems!

Note by Daniel Liu
5 years, 11 months ago

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Comments

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There, done with my first #CosinesGroup post. Feel free to give any feedback!

I hope all of you can learn something from this, if you haven't learned it already. :)

Daniel Liu - 5 years, 11 months ago

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This post is amazingreally big number\text{amazing}^{\text{really big number}}.

Way to go! Please, do more! I'd like to take part in too.:D

Congratulations!

Guilherme Dela Corte - 5 years, 11 months ago

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Thank you! If you want to make a #CosinesGroup post, refer to this thread: https://brilliant.org/discussions/thread/math-circles/

Daniel Liu - 5 years, 11 months ago

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One of the best question i have ever seen!!! Thank you!!!!!!

Sharath Chandar - 5 years, 11 months ago

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You're very welcome. Glad I could help :)

Daniel Liu - 5 years, 11 months ago

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This is very well-written. Good job!

Mursalin Habib - 5 years, 11 months ago

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Impressive! Also, very well explained, thank you!

Wesley Khalil - 5 years, 11 months ago

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good work

rajasekhar e - 5 years, 11 months ago

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Well Explained... Thank You !! :)

paramesh king - 5 years, 11 months ago

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That is really nice! Thank you very much.

Toan Pham Quang - 5 years, 11 months ago

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very much interesting... Thanks for enlighten me..

Ravishankar Shukla - 5 years, 11 months ago

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Nice explanation

Ben Habeahan - 5 years, 11 months ago

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Nice and very well-explained post!

Muh. Amin Widyatama - 5 years, 11 months ago

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Wow, It's really great and helpful. Thanks

Rahat Khan - 5 years, 11 months ago

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cool, this one is my quiz in solid mensuration, could you make like that explanation in analytic geometry, cause i'm in trouble in that subject.

Jeriel Villa - 5 years, 11 months ago

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Not really sure what you mean by "solid mensuration"... What exactly do you want me to do? Some more analytic geometry posts?

Daniel Liu - 5 years, 11 months ago

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Very nice! Only now I'm thinking... What about the volume between two concentric spheres, with a disk that is tangent to the smaller sphere and whose circular boundary is on the larger sphere...

Matt Enlow - 5 years, 11 months ago

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There is a 3-D analogue of this property. I'm quite certain that you have seen it before, especially if I were to mention it.

Calvin Lin Staff - 5 years, 11 months ago

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can i get the solution for problem 3 ?

Shujaat Khan - 5 years, 11 months ago

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@Shujaat Khan Note that a side length of each polygon acts as a chord. Since we know that the area between the two circles are based only on the chord's length, and since the lengths of the two chords in both cases are the same, the area ratio must be 1:1.

Daniel Liu - 5 years, 11 months ago

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@Daniel Liu thanks for clearing the question.

Shujaat Khan - 5 years, 10 months ago

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@Shujaat Khan 1:4

Basavaraj S Mat - 5 years, 11 months ago

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1) 25 \pi 2)8 \sqrt{2} 3)1:1 4)length of half chord= \sqrt{R^{2}-r^{2}}= x/2 \pi (R^{2}-r^{2})= (x^2 \pi)/4

Ayush Alankar - 5 years, 11 months ago

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Correct!

Daniel Liu - 5 years, 11 months ago

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Good job, thanks for sharing

Rafael Dias - 5 years, 11 months ago

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Amazing

Rafael Dias - 5 years, 11 months ago

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Answer for question 2 is squareroot(128).Is it right?

Basavaraj S Mat - 5 years, 11 months ago

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Partially correct. You forgot to reduce to simplest radical form ;)

Daniel Liu - 5 years, 11 months ago

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I loved ur methodology , keep posting such things

Aayush Chaturvedi - 5 years, 11 months ago

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You are a nice guy and I liked this note \infty^{\infty}

Soham Dibyachintan - 5 years, 11 months ago

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Surely, an upvote for it But, the interesting part is that I got the exactly same question in my quiz that I attempted 3 years ago. I still remember it, because this was the most interesting question for which I successfully found a solution

Amazed to see the same question again after 3 years.

Kishlaya Jaiswal - 5 years, 11 months ago

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Good job! You got good skills! :)

Akshat Jain - 5 years, 11 months ago

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Tried it a different more complicated way (found the radii) and got 16 pi. Your way is so much better!! Good job my friend! :)

Scholastica Okoye - 5 years, 10 months ago

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please do more of these!!!!!!!!!!!!! ;)

Nishchay Selot - 5 years, 10 months ago

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Very well explained, thanks!

Neelansh Bute - 5 years, 10 months ago

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nice

Mary Love Hempiso - 5 years, 10 months ago

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@Daniel Liu loved it :)

Abhinav Raichur - 5 years, 1 month ago

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Outstanding note.

Swapnil Das - 4 years, 8 months ago

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2+3

Akbar Shah - 5 years, 11 months ago

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I have already learned this in high school

Niraj Sapkota - 5 years, 11 months ago

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