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We have the circle $(x-a)^2 + (y-a)^2 = a^2$ (which always is tangent to both of the axes). There are 4 of these circles which are tangent to the circle $x^2+y^2=2$. Get the 2 positive values for $a$ at which the circles are tangent (make a sketch to find the point of tangency).

Can anyone help me with this, I have no clue how to exactly get the values of $a$.

JohnPhteven
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    Did you follow the suggestion to make a sketch? What did you learn from it? – Harald Hanche-Olsen Nov 01 '12 at 16:54
  • I did, it didn't really help me – JohnPhteven Nov 01 '12 at 16:59
  • What do you know about the center of the circle $(x-a)^2 + (y-a)^2 = a^2$ and it's radius? You want two positive values of $a$ such that the corresponding circles touch the circle $x^2+y^2=2$ (for which the center is ___ and radius ____). Now, you need to clarify whether the "touching" means tangency (one point in common) and/or intersection (overlapping, with two points in common). – amWhy Nov 01 '12 at 17:01
  • It means tangency, not intersection. – JohnPhteven Nov 01 '12 at 17:04
  • If you understand the answer to this question, you should be able to draw and make sense of the sketch - which has a natural axis of symmetry and on which you should mark the centres and radii of your circles. Have you managed to create separate sketches for the four circles mentioned (now an answer has been given)? Because a good sketch makes this kind of question a whole lot easier for most people. – Mark Bennet Nov 01 '12 at 17:59

1 Answers1

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To touch externally, the distance between the centres = the sum of the radii.

So, $(a-0)^2+(a-0)^2=(|a|+2)^2$

(i)$\implies 2a^2=a^2+4|a|+4\implies a^2-4|a|-4=0$

As $a\ge0,a^2-4a-4=0, a=2\pm2\sqrt2$

As $a\ge0,a=2+2\sqrt2$

alternatively,(ii) $2a^2=(a+2)^2\implies \sqrt2a=a+2 $ as $a>0,a+2>0$ $\implies(\sqrt2-1)a=2 \implies a=\frac 2{\sqrt2-1}=2(\sqrt2 +1)$

To touch internally, the distance between the centres = the difference of the radii.

So, $(a-0)^2+(a-0)^2=(|a|-2)^2$

(i)$a^2+4|a|-4=0$

As $a>0,a^2+4a-4=0,a=-2\pm2\sqrt 2$

So, $a=2\sqrt2-2$

alternatively,(ii) $2a^2=(a-2)^2$

$\implies \sqrt2a=a-2$ if $a\ge 2$

$\implies (\sqrt2-1)a=-2\implies a=-\frac{2}{\sqrt 2-1}<0<2$ which is impossible.

$\implies \sqrt2a=2-a$ if $a< 2$

$\implies a(\sqrt2+1)=2\implies a=2(\sqrt 2-1)$

lab bhattacharjee
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