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18 Mar 2019, 02:53
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
29% (03:16) correct
71%
(02:44)
wrong
based on 38
sessions
History
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A circle centered at A with a radius of 1 and a circle centered at B with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. The radius of the third circle is
(A) 1/3
(B) 2/5
(C) 5/12
(D) 4/9
(E) 1/2
Attachment:
e0340e3581aab9d635bcf2dca6b71152f8fa5290.png [ 20.01 KiB | Viewed 5354 times ]
01 Apr 2019, 04:10
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Bunuel
It can be solved by calculating the length of common tangent for all three circles
In the attached picture
Let the third circle be denoted by C and let the radius be x
common tangent LN = LM+MN
lenght of LN = √((distance between centres of circle A and B )^2 -( difference of radius of circle A and B))
= √(4+1)^2+(4-1)^2
= √25-9
=√16
= 4
Similarly we calculate LM and MN by using the above mentioned formula
now we have ,
LN = LM+MN
after putting the values of LM and MN and LN we get the below mentioned equation
4 = √(4x) +√(16x)
4 = 2√x+4√x
4 = 6√x
means x = 4/9
Award kudos if helpful ..
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