The radius of the larger circle is $10$ cm.

Find the radius of the largest circle that will fit in the middle.


From my IGCSE math textbook.

I have tried to solve it but its too hard for me.

First I tried making a square by joining the smaller diameters.One diameter plus the distance from it to to centre=x. The answer is $\sqrt{2x^2}$ But how to find x, I don't know.

Can someone come up with a solution that's understandable and suitable for me? I am an IGCSE (Extended) student.

  • 2
    $\begingroup$ Why is it too hard? What have you tried to solve it already? Edit your question to add in these details. People will be more willing to answer your questions if they can see you have put in some effort yourself first $\endgroup$ – lioness99a Mar 13 at 9:50
  • $\begingroup$ Hint: connect the radius of the 4 inner circles. $\endgroup$ – user690234 Mar 13 at 9:50
  • $\begingroup$ Tried putting lines around it and and trying Pythagoras Theorem, but the length from center to an imaginary bounding square's corner needs to be found out. I wiil edit right now $\endgroup$ – 4R1u Mar 13 at 9:53
  • $\begingroup$ Also, have a read of this thread which asks the same question $\endgroup$ – lioness99a Mar 13 at 9:53
  • $\begingroup$ Yes, I put the image link in first version from there $\endgroup$ – 4R1u Mar 13 at 10:12

Consider the square formed by the centers of the four medium circles. Its side is the diameter of the medium circles.

The diameters of the large and small circles are the diagonal of the square, plus or minus the diameter of a medium circle.

Hence the small diameter


enter image description here

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  • $\begingroup$ Thanks for the simplification. $\endgroup$ – 4R1u Mar 13 at 11:02

Denote the diameters $D_2 < D_1 < D_0$. We immediately find $$D_0 = 2D_1 + D_2.$$ Moreover, connecting the centers of the four smaller circles yields a square with side length $D_1$. Thus we can compute its diagonal by $$\sqrt{2D_1^2} = D_1 + D_2.$$ Solving for $D_1$ we get $$D_1 = \dfrac{1}{\sqrt{2}-1} D_2.$$ Plugging this into the first equation lets you compute the desired value $D_2$.

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  • $\begingroup$ I really appreciate your answer, it is elaborated than my selected one.This is easier to understand $\endgroup$ – 4R1u Mar 13 at 11:03
  • $\begingroup$ You are welcome. I think the answers are basically the same and I learned alot by answering it (I would not have thought that the inner circle is the way to compute the four other ones...) $\endgroup$ – PrudiiArca Mar 13 at 11:08

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