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Two circles of radius $8$ are placed inside a semi-circle of radius $25$.The two circles are each tangent to the diameter and to the semi-circle. If the distance between the centers of the two circles is $\lambda$, then the value of $\frac{60}{\lambda}$.

My try: I am not sure if I am correct, I guess these two circles touch each other and $\lambda$ is $16$.Can someone explain me correct answer and correct method.

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  • $\begingroup$ The question did not say they are touching each other. Therefore, $\lamda$ is not 16. $\endgroup$
    – Mick
    Jul 29 '15 at 8:09
  • $\begingroup$ @Mick,yes question does not say they touch each other. $\endgroup$ Jul 29 '15 at 8:10
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    $\begingroup$ I am sorry, what is the question? $\endgroup$
    – nicks
    Jul 29 '15 at 9:20
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Try to figure that out from the attached diagram.

enter image description here

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  • $\begingroup$ ,you drew this diagram in GeoGebra? $\endgroup$ Jul 29 '15 at 8:27
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    $\begingroup$ Definitely a +1 for the diagram. A pic is worth a thousand words. I had a hard time describing my answer without being able to draw. $\endgroup$
    – Shailesh
    Jul 29 '15 at 8:33
  • $\begingroup$ @VinodKumarPunia Yes. It is a great tool. $\endgroup$
    – Mick
    Jul 29 '15 at 11:37
  • $\begingroup$ @Shailesh In the field of Geogebra, I am a newbie also because many useful applications are still "hidden" (to me). $\endgroup$
    – Mick
    Jul 29 '15 at 11:41
  • $\begingroup$ @Mick. Good job. Only point is to actually show that the line through centers also passes through the point of tangency. $\endgroup$
    – Shailesh
    Jul 29 '15 at 13:42
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Considering a line from the centre of the semicircle through the centre of one of the circles, the distance from the centre of the semicircle to the centre of the small circle is $25-8=17$. Then applying pythagoras gives the distance between the centres $=\lambda=2\times15=30$

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Hint :

From the midpoint of the larger semicircle, draw a line through the center of one of the smaller semi-circle, extending upto the the larger semi-circle. This is the point of tangency. Why ? . Now, let this line make an angle $\theta$, set up equations, and you will find a nice 8-15-17 right triangle. So the 'horizontal' length between the centers is 15. Twice that is 30. So $\lambda$ = 2. I can see that David Quinn has given a similar answer.

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Consider the segment with endpoints at the center of the semicircle and at the center of one of the circles. It's length is $25-8=17$. Now, think of a right triangle whit this segment as hypotenuse and with one side upon the diameter. This side has length 15, because of the pythagorean theorem and the fact that the other sides are 8 and 17. The same is true for the other circle and if we add the horizontal distances (consider the diameter of the semicircle is horizontal) we get that $\lambda = 30$

Edit: later i saw that other people posted similar answers

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