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# help

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In the Figure ,

when all the outer circle have radii r,

then the radius of the inner circle is what, in terms of r? Jun 14, 2020

#1
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If you connect the radii of each circle to its tangent points with the other circles, you have a square with side lenth 2r.

The diagonal of this square is $$2r\sqrt{2}$$ by the pythagorean theorem.

The diagonal is r+small circle diameter+r = $$2r\sqrt{2}$$

Subtracting both sides by 2r, the diameter of the samll circle should be $$2r\sqrt{2}-2r$$ which means the radius should be $$\frac{2r\sqrt{2}-2r}{2}=\sqrt{2}r-r=r(\sqrt{2}-1)$$.

Jun 14, 2020
edited by MIRB16  Jun 14, 2020
edited by MIRB16  Jun 14, 2020
#2
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Let the inner circle's radius be x. We are solving for x in terms of r.

So connect the centers of two adjacent circles with radii r and the center of the small circle with radius x.

This is a isoceles right triangle with sides r+x, r+x, and 2r.

We know that through the Pythagorean Theorem, the ratio of a side to the hypotenuse is $$1:\sqrt 2$$.

So $$\sqrt 2(r+x)=2r$$. Divide both sides by $$\sqrt 2$$ and rationalize to get $$r+x=\sqrt 2 r$$. Minus both sides by r and factor out r to get our answer: $$x = (\sqrt 2-1)r$$

Jun 14, 2020