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A circle with radius 2 is placed against a right angle. Another small circle is also placed as shown in the adjoining figure. What is the radius of the smaller circle?

1. $3-2 \sqrt{2}$
2. $4-2 \sqrt{2}$
3. $7-4 \sqrt{2}$
4. $6-4 \sqrt{2}$

If we draw a line joining the centres of the two circles and leading to the intersection point of the two tangents , that can help us arrive at the solution.

Let r be the radius of the smaller circle so distance between centre and the point of intersection of tangent = sqrt(r2 + r2)   =  √2 r.

And similarly distance between centre of larger circle and the point of intersection of tangent  =  √ (2+ 22)  =  2√2 r

Also this distance can be written as : radius of larger circle + radius of smaller circle + distance between centre of smaller circle to point of intersection of tangents              =    2 + r + √2 r.

Hence we have :

2 + r + √2 r     =     2√2

==>    r(1 + √2)         =     2(√2 - 1)

==>    r                    =     2(√2 - 1) / (√2 + 1)

==>    r                    =   [  2(√2 - 1) / (√2 + 1)  ]  * [ (√2 - 1) / (√2 - 1) ]

=      2(√2 - 1)2 / [ (√2)2 -  (1)2 ]

=      2(√2 - 1)2

=      6 - 4√2

Hence 4) should be the correct answer.