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 \(8\), 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.

First, denote the center to be \(O\). Let the point of tangency of the chord on the little circle be \(A\), and an endpoint of the chord be \(B\). Finally, let the radius of the small circle be \(r\) and the radius of the large circle be \(R\).

Notice that \(AB=4\), because \(AB\) is half of the chord, which has length \(8\). Since \(\angle OAB=90^{\circ}\) because \(AB\) is tangent to the small circle, we can use the Pythagorean Theorem on \(\Delta OAB\).

Using it gives \(r^2+4^2=R^2\), or \(16=R^2-r^2\). But now it seems like we're stuck. Not only do we not know \(r\), but we also do not know \(R\). 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 \(\pi R^2\), and the area of the small circle is \(\pi r^2\); therefore the area in between is \(\pi R^2-\pi r^2\).

But wait a minute! \(\pi R^2-\pi r^2=\pi (R^2-r^2)\), and we know the value of \(R^2-r^2=16\)! So therefore the area in between the two circles is \(\pi (16)=\boxed{16\pi}\) 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 \(r\) into anything you want: \(3\), \(2014\), even \(\sqrt{2}+\pi\); and see if the area in between changes or not.

A special case of \(r\) might have come up to your mind now: *what if* \(r=0\)*?*

In that case, the circle in the center disappears, and the problem degenerates to finding the area of a circle with diameter \(8\)! And when we find the area, we see that indeed, the area still equals \(16\pi\). 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 \(10\). What is the area in between the two circles?

**2.** Three concentric circles \(A,B,C\) have the property that the radius of \(A\) is bigger than the radius of \(B\), and the radius of \(B\) is bigger than the radius of \(C\). A chord of circle \(B\) of length \(8\) is drawn such that it is tangent to circle \(C\); a chord of circle \(A\) of length \(x\) is drawn such that it is tangent to circle \(B\). If the area between circle \(A\) and \(B\) is twice the area between circle \(B\) and circle \(C\), then what is the value of \(x\)? 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 \(A\). A regular decagon is also both inscribed and circumscribed by a circle. Let the area between those circles be \(B\). If the side lengths of the two polygons are equal, then what is the ratio of \(A:B\)?

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

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

## Comments

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TopNewestThere, 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 · 2 years, 10 months ago

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

Congratulations! – Guilherme Dela Corte · 2 years, 10 months ago

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– Daniel Liu · 2 years, 10 months ago

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

One of the best question i have ever seen!!! Thank you!!!!!! – Sharath Chandar · 2 years, 10 months ago

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– Daniel Liu · 2 years, 10 months ago

You're very welcome. Glad I could help :)Log in to reply

This is very well-written. Good job! – Mursalin Habib · 2 years, 10 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 · 2 years, 10 months ago

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– Daniel Liu · 2 years, 10 months ago

Correct!Log in to reply

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 · 2 years, 10 months ago

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– Calvin Lin Staff · 2 years, 10 months ago

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.Log in to reply

– Shujaat Khan · 2 years, 10 months ago

can i get the solution for problem 3 ?Log in to reply

– Daniel Liu · 2 years, 10 months ago

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.Log in to reply

– Shujaat Khan · 2 years, 9 months ago

thanks for clearing the question.Log in to reply

– Basavaraj S Mat · 2 years, 10 months ago

1:4Log in to reply

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 · 2 years, 10 months ago

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– Daniel Liu · 2 years, 10 months ago

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

Wow, It's really great and helpful. Thanks – Rahat Khan · 2 years, 10 months ago

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Nice and very well-explained post! – Muh. Amin Widyatama · 2 years, 10 months ago

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Nice explanation – Ben Habeahan · 2 years, 10 months ago

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very much interesting... Thanks for enlighten me.. – Ravishankar Shukla · 2 years, 10 months ago

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That is really nice! Thank you very much. – Toan Pham Quang · 2 years, 10 months ago

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Well Explained... Thank You !! :) – Paramesh King · 2 years, 10 months ago

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good work – Rajasekhar E · 2 years, 10 months ago

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Impressive! Also, very well explained, thank you! – Wesley Khalil · 2 years, 10 months ago

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Outstanding note. – Swapnil Das · 1 year, 7 months ago

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@Daniel Liu loved it :) – Abhinav Raichur · 2 years ago

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nice – Mary Love Hempiso · 2 years, 9 months ago

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Very well explained, thanks! – Neelansh Bute · 2 years, 9 months ago

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please do more of these!!!!!!!!!!!!! ;) – Nishchay Selot · 2 years, 10 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 · 2 years, 10 months ago

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Good job! You got good skills! :) – Akshat Jain · 2 years, 10 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 · 2 years, 10 months ago

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You are a nice guy and I liked this note \[\infty^{\infty}\] – Soham Dibyachintan · 2 years, 10 months ago

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I loved ur methodology , keep posting such things – Aayush Chaturvedi · 2 years, 10 months ago

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Answer for question 2 is squareroot(128).Is it right? – Basavaraj S Mat · 2 years, 10 months ago

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– Daniel Liu · 2 years, 10 months ago

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

Amazing – Rafael Dias · 2 years, 10 months ago

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Good job, thanks for sharing – Rafael Dias · 2 years, 10 months ago

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2+3 – Akbar Shah · 2 years, 10 months ago

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I have already learned this in high school – Niraj Sapkota · 2 years, 10 months ago

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