My Math Forum Circle Problem

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 September 19th, 2015, 07:55 PM #1 Member     Joined: Sep 2013 From: New Delhi Posts: 38 Thanks: 0 Math Focus: Calculus Circle Problem If the diameters intersect at 90°, the radius of bigger circle is 1 m, what is the radius of smaller circle ?
 September 20th, 2015, 03:21 AM #2 Newbie   Joined: Sep 2015 From: Uk Posts: 4 Thanks: 1 Thanks from aditya ranjan
 September 20th, 2015, 05:44 AM #3 Senior Member     Joined: Feb 2010 Posts: 703 Thanks: 138 $\displaystyle r \sqrt{2} = 1 - r$ $\displaystyle r = -1 + \sqrt{2}$ Thanks from aditya ranjan and aurel5
September 21st, 2015, 08:57 AM   #4
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Math Focus: Calculus
Quote:
 Originally Posted by mrtwhs $\displaystyle r \sqrt{2} = 1 - r$ $\displaystyle r = -1 + \sqrt{2}$
I understood Jtatezen's method but how did you solve it?

 September 21st, 2015, 09:52 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1122 Math Focus: Elementary mathematics and beyond $\displaystyle \sqrt2r+r=1,\quad r=\dfrac{1}{\sqrt2+1}=\sqrt2-1$
September 21st, 2015, 04:06 PM   #6
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Quote:
 Originally Posted by aditya ranjan 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.

 September 21st, 2015, 05:18 PM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1122 Math Focus: Elementary mathematics and beyond 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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