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

Are You Up For the Challenge: 700 Level Questions

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.

Posted from my mobile device

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)

Posted from my mobile device

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 6893 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

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards