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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 $a$.

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    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.2012-11-01

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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 $a< 2$

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

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    Thank you very much, you have done$a$terrific job! I understand it.2012-11-01