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  
September 20th, 2015, 05:44 AM  #3 
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 140 
$\displaystyle r \sqrt{2} = 1  r$ $\displaystyle r = 1 + \sqrt{2}$ 
September 21st, 2015, 08:57 AM  #4 
Member Joined: Sep 2013 From: New Delhi Posts: 38 Thanks: 0 Math Focus: Calculus  
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 
$\displaystyle \sqrt2r+r=1,\quad r=\dfrac{1}{\sqrt2+1}=\sqrt21$

September 21st, 2015, 04:06 PM  #6 
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 140  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 $\displaystyle 1r$. 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): $\displaystyle A=\dfrac{\sqrt2}{2}r+\dfrac{\sqrt2}{2}r+r=1 \Rightarrow r(\sqrt2+1)=1\Rightarrow r=\sqrt21$ 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 onehalf the base times the height. 

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