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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: 706 Thanks: 140 $\displaystyle r \sqrt{2} = 1 - r$ $\displaystyle r = -1 + \sqrt{2}$ Thanks from aditya ranjan and aurel5 September 21st, 2015, 08:57 AM   #4
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 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: 7,935 Thanks: 1129 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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 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: 7,935 Thanks: 1129 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. Tags circle, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post kaspis245 Geometry 3 August 9th, 2014 11:37 AM shiseonji Algebra 10 April 17th, 2014 03:43 PM techamitdev Algebra 1 August 8th, 2009 05:13 AM sujitkb Algebra 2 March 3rd, 2009 10:01 PM planke Algebra 1 November 15th, 2007 03:14 AM

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