in Quantitative Aptitude edited by
2,363 views
1 vote
1 vote

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. $6-4 \sqrt{2}$
in Quantitative Aptitude edited by
13.4k points
2.4k views

1 Answer

0 votes
0 votes
Best answer

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.

selected by
1.5k points

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true