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September 19th, 2015, 07:55 PM   #1
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Exclamation Circle Problem

If the diameters intersect at 90°, the radius of bigger circle is 1 m, what is the radius of smaller circle ?

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September 20th, 2015, 03:21 AM   #2
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September 20th, 2015, 05:44 AM   #3
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$\displaystyle r \sqrt{2} = 1 - r$

$\displaystyle r = -1 + \sqrt{2}$
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September 21st, 2015, 08:57 AM   #4
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Originally Posted by mrtwhs View Post
$\displaystyle r \sqrt{2} = 1 - r$

$\displaystyle r = -1 + \sqrt{2}$
I understood Jtatezen's method but how did you solve it?
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September 21st, 2015, 09:52 AM   #5
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$\displaystyle \sqrt2r+r=1,\quad r=\dfrac{1}{\sqrt2+1}=\sqrt2-1$
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September 21st, 2015, 04:06 PM   #6
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Originally Posted by aditya ranjan View Post
I understood Jtatezen's method but how did you solve it?
Let r be the radius of the small circle. Draw a radius of the large circle that also goes through the center of the small circle. The distance between the centers is $\displaystyle r \sqrt{2}$. It is also $\displaystyle 1-r$. Now solve it using greg1313's method.
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September 21st, 2015, 05:18 PM   #7
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Alternatively, extend the given radii an equal distance, so that the segment between their endpoints is tangent to the circle. This gives a triangle with sides $\displaystyle \sqrt2,\sqrt2$ and $\displaystyle 2$ (use the Pythagorean theorem). The area of this triangle is $\displaystyle 1$, but we may also calculate the area using the radius of the incircle (which is the radius we want to find):

$\displaystyle A=\dfrac{\sqrt2}{2}r+\dfrac{\sqrt2}{2}r+r=1 \Rightarrow r(\sqrt2+1)=1\Rightarrow r=\sqrt2-1$

This uses the fact that the radii are perpendicular to the sides of the triangle where the incircle is tangent to the triangle. Recall that the area of a triangle may be computed by multiplying one-half the base times the height.
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