Last visit was: 24 Apr 2026, 04:16 It is currently 24 Apr 2026, 04:16
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
705-805 (Hard)|   Geometry|      
User avatar
v12345
User avatar
Joined: 01 Mar 2015
Last visit: 19 Jan 2026
Posts: 398
Own Kudos:
1,117
 [6]
Given Kudos: 44
Location: India
Posts: 398
Kudos: 1,117
 [6]
Three circles of radius 20 are arranged with their respective centers 17 Sep 2021, 03:19
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

85% (hard)

Question Stats:

41% (02:23) correct 59% (02:37) wrong based on 41 sessions

History

Date
Time
Result
Not Attempted Yet
Three circles each of radius 20 are arranged with their respective centers A, B and C in a row. If the line WZ is tangent to the third circle, find the length of XY.
Attachment:
Capture.PNG
Capture.PNG [ 37.23 KiB | Viewed 3449 times ]
A. 28
B. 30
C. 32
D. 34
E. 36
ShowHide Answer
Official Answer
C
User avatar
v12345
User avatar
Joined: 01 Mar 2015
Last visit: 19 Jan 2026
Posts: 398
Own Kudos:
1,117
 [1]
Given Kudos: 44
Location: India
Posts: 398
Kudos: 1,117
 [1]
Three circles of radius 20 are arranged with their respective centers 17 Sep 2021, 11:52
Kudos
Add Kudos
Bookmarks
Bookmark this Post
r = 20 (given)
Let P be the midpoint of XY and Q be the point where the line WZ meets the third circle.
Attachment:
image.PNG
image.PNG [ 19.23 KiB | Viewed 3365 times ]
Then by similar triangles WPB and WQC,
\(\frac{PB}{WB} = \frac{QC}{WC}\)
\(\frac{PB}{3r} = \frac{r}{5r}\)
\(\frac{PB}{60} = \frac{1}{5}\)
\(PB = 12\)

\(XP = \sqrt{XB^2 - PB^2} = \sqrt{20^2 - 12^2} = 16\)
XY = 2*XP = 32
Hence, OA is (C).
avatar
piyushyadav9801
avatar
Joined: 11 Apr 2021
Last visit: 12 May 2024
Posts: 1
Given Kudos: 4
Posts: 1
Kudos: 0
Three circles of radius 20 are arranged with their respective centers 18 Sep 2021, 04:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
v12345
r = 20 (given)
Let P be the midpoint of XY and Q be the point where the line WZ meets the third circle.
Attachment:
image.PNG
Then by similar triangles WPB and WQC,
\(\frac{PB}{WB} = \frac{QC}{WC}\)
\(\frac{PB}{3r} = \frac{r}{5r}\)
\(\frac{PB}{60} = \frac{1}{5}\)
\(PB = 12\)

\(XP = \sqrt{XB^2 - PB^2} = \sqrt{20^2 - 12^2} = 16\)
XY = 2*XP = 32
Hence, OA is (C).
Show more

In the last step, how have we considered Triangle XPB to be a right angle triangle? Please explain.
User avatar
Fdambro294
User avatar
Joined: 10 Jul 2019
Last visit: 20 Aug 2025
Posts: 1,331
Own Kudos:
772
Given Kudos: 1,656
Posts: 1,331
Kudos: 772
Three circles of radius 20 are arranged with their respective centers 02 Oct 2021, 22:44
Kudos
Add Kudos
Bookmarks
Bookmark this Post
You assume that the line segment is perpendicular to that long hypotenuse.

Then you can use the rule that any line drawn from the center of a circle that is Perpendicular to a chord will BISECT that chord.

Thus, P will Bisect the chord XY because line BP is assumed to be perpendicular to that chord.

You find the ratio of the similiar triangles as 5 : 3

Which would then make the corresponding sides ———> (Radius from center C = 20) and BP in that same ratio.

5/3 = 20/BP

BP = 12

Then draw a hypotenuse/radius from center B to point X

You have right triangle BPX with leg of 12 and hypotenuse of 20

This is a 3-4-5 right triangle.

XP = 16

And again, referring back to the earlier rule that a line drawn from the center of a circle Perpendicular to a chord Bisects that chord

XP = PY = 16

Thus XY = (2) (16) =

32

Hopefully some of it was helpful?

The key is to assume that BP is perpendicular to the long line WZ, and then utilizing the Rule that any line drawn from the center of a circle Perpendicular to a chord will Bisect that chord.


piyushyadav9801
v12345
r = 20 (given)
Let P be the midpoint of XY and Q be the point where the line WZ meets the third circle.
Attachment:
image.PNG
Then by similar triangles WPB and WQC,
\(\frac{PB}{WB} = \frac{QC}{WC}\)
\(\frac{PB}{3r} = \frac{r}{5r}\)
\(\frac{PB}{60} = \frac{1}{5}\)
\(PB = 12\)

\(XP = \sqrt{XB^2 - PB^2} = \sqrt{20^2 - 12^2} = 16\)
XY = 2*XP = 32
Hence, OA is (C).

In the last step, how have we considered Triangle XPB to be a right angle triangle? Please explain.
Show more

Posted from my mobile device
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,967
Own Kudos:
1,117
Posts: 38,967
Kudos: 1,117
Three circles of radius 20 are arranged with their respective centers 07 Aug 2025, 07:14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
Moderators:
Bunuel
Math Expert
109809 posts
Krunaal
Tuck School Moderator
853 posts