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PostPosted: Sun, 5 Aug 2007 22:03:42 UTC 
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Question: Two different circles that passes through the point C(1,3) are tangent to both coordinate axes. If the length of a radius of the smaller circle is r and the length of a radius of the larger circle is R, what is the value of r+R?
Code:
Diagram:
         
          o  o
 y|    o        o
  |   
  | o              o
  |
  |o        B       o     where B marks the center of the larger circle
  |                             C marks the point of intersection of the two circles
  | C* *           o            A marks the center of the smaller circle
  |*  A  *
  |*   o *      o
 _| _*_*__o__o_________
 O|
Spoiler:
The line that connects the centers of the two circles A(r,r) and B(R,R) is y=x.
The slope of the line that connects the origin O and C(1,3) is y=3x.
Let the angle (<COA = <COB) between y=3x and y=x be \theta.
Then \displaystyle\tan\left(\theta+\frac{\pi}{4}\right)=\frac{\tan\theta +1}{1-\tan\theta}=3\Rightarrow\tan\theta=\frac{1}{2}
Therefore, \displaystyle\cos\theta=\frac{2}{\sqrt{5}}
Code:
Diagram
  |        . B
  |     .   /
  |   C   /
  |   .\
  |  .  A
  | . /
 _|./____ _
 O|
Looking at triangle \Delta COA and triangle \Delta COB, \overline{OB}=\sqrt{1^2+3^2}=\sqrt{10}, \overline{OA}=r\sqrt{2} and \overline{OB}=R\sqrt{2}.
For \Delta COA, by cosine law, r^2=\overline{OC}^2+\overline{OA}^2-2(\overline{OC})(\overline{OA})\cos\theta
We have, r^2=10+2r^2-2(\sqrt{10})(r\sqrt{2})(\frac{2}{\sqrt{5}}) or r^2-8r+10=0...[1]
Similarly, for \Delta COB, R^2=\overline{OC}^2+\overline{OB}^2-2(\overline{OC})(\overline{OB})\cos\theta
We have, R^2=10+2R^2-2(\sqrt{10})(R\sqrt{2})(\frac{2}{\sqrt{5}}) or R^2-8R+10=0...[2]
The roots of the two equations for r and R are the same. That implies the smaller root equals r and the larger root equals R. Therefore the sum r+R equals the sum of the roots which is given by \displaystyle r+R=-\frac{b}{a}=8

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PostPosted: Sun, 5 Aug 2007 23:36:08 UTC 
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Joined: Fri, 20 Jul 2007 22:50:53 UTC
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Location: Texas
Spoiler:
Half of that proof seems unnecessary to me as only 2 circles tangent to both the x-axis and y-axis in the first quadrant will intersect a point (not on the axis) in quadrant I in the first place.

(1-r)^{2}+(3-r)^{2}=r^{2}

r^{2}-8r=-10
(r-4)^{2}=6
r=4\pm\sqrt{6}

r_1+r_2=4+\sqrt{6}+4-\sqrt{6}=8


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 Post subject: Answer
PostPosted: Thu, 25 Sep 2008 08:42:52 UTC 
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Joined: Wed, 24 Sep 2008 12:44:04 UTC
Posts: 5
First observe that the coordinates of A and B are respectively (r,r) and (R,R), therefore the equations of the circles are given by (x-t)^2+(y-t)^2=t^{2} with t=r for the small circle, etc. Now use the fact that C=(1,3) belongs to both of them, then we get
t^2-8t+10= for t=r,R. Thus R^2-r^2-8(R-r)=0 and it follows that R+r=8.


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