Two circles, each of radius 4 cm, touch externally. Each of these two

Question

Competition Mode Question

Two circles, each of radius 4 cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then what is the radius of the third circle?

A. 1/√2

B. 1

C. π/3

D. √2

E. √3

A.  \frac{1}{√2}

B.  1

C.  \frac{π}{3}

D.  √2

E.  √3

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GMATinsight · Accepted Answer

Please find attached the video solution. It's a hard question so steps would be really useful to understand the approach.

The question has been answered without preparation in the first attempt.

Answer: Option B

https://www.youtube.com/watch?v=BZYMkesQuUo

yashikaaggarwal · Answer

Radius of the smaller circle is 1. You can get this by using Pythagoras theorem.
Draw a line from the centre of smaller circle to that of larger circle such that it will form hypotenuse of the triangle. Which will be (4+r) =radius of big circle+ radius of smaller circle.
Base is 4 
And height be H
H will be √r^2+8r
Then the height till the tangent is h+r which is 4
Equating that will give
(√r^2+8r)+r=4
√r^2+8r=4-r
Squaring both sides
r^2+8r=r^2+16-8r
16r=16
R=1

So OA is B.

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freedom128 · Answer

From the attached sketch, we can solve following equation:  4^2 + x^2 = (4+r)^2 , where  r  is the radius of third circle

Since  x=4-r ,
4^2 + (4-r)^2 = (4+r)^2
16 + 16 -8r +r^2 = 16 +8r +r^2
16=16r
 r=1

FINAL ANSWER IS (B)

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unraveled · Answer

Two circles, each of radius 4 cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then what is the radius of the third circle?

A. \(\frac{1}{√2}\)

B. 1

C. \(\frac{π}{3}\)

D. √2

E. √3

Please refer snapshot.

Attachment:

Two circles 4cm.png [ 23.66 KiB | Viewed 7011 times ]

Let the radius of the third circle be r.
From the figure using Pythagoras theorem
\((4+r)^2 = 4^2 + (4-r)^2\)
\(16 + 8r + r^2 = 16 + 16 - 8r + r^2\)
r = 1
Answer B.

exc4libur · Answer

note: the only way this works is if the third circle is very small and placed between the other equal circles

distance from the large circle to the tangent is 4
distance from the small circle to the tangent is r
imagine a triangle with its vertices equal to the origin of each circle
base will have 4+4=8
height will have 4-r
other sides will have 4+r
if we bisect the triangle from its height we get a right triangle
using Pythagoras we have:
4^2+(4-r)^2=(4+r)^2
16=(4+r)^2-(4-r)^2
16=8r-(-8r)
r=1

Ans (B)

NutHead · Answer

Unable to attach photo. so just writing the equation here.

Let the smaller circle be of r radius, then -> (4+r)^2 = (4-r)^2 + 4^2 => r=1
Answer B

razorhell · Answer

Please refer to attached image ..

Let the radius be r

Using pytharos we will get :

(4+r)^2 = 4^2 + (4-r)^2

Solving we get r =1

Ans B

ScottTargetTestPrep · Answer

bumpbot · Answer

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Two circles, each of radius 4 cm, touch externally. Each of these two

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Two circles, each of radius 4 cm, touch externally. Each of these two

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