Last visit was: 04 Jun 2026, 01:23 It is currently 04 Jun 2026, 01:23
Home
Messages
MBA Fair
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
Sub 505 (Easy)|   Geometry|      
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 04 Jun 2026
Posts: 111,096
Own Kudos:
818,718
 [4]
Given Kudos: 106,641
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 111,096
Kudos: 818,718
 [4]
The circle with center C shown above is tangent to both axes and has 22 Aug 2019, 01:19
1
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

5% (low)

Question Stats:

89% (01:41) correct 11% (01:59) wrong based on 133 sessions

History

Date
Time
Result
Not Attempted Yet

The circle with center C shown above is tangent to both axes and has an area of A. What is the distance from O to C, in terms of A?

A. A

B. \(\sqrt{\frac{2A}{\pi}}\)

C. \(\sqrt{\frac{3A}{\pi}}\)

D. \(\frac{2A}{\pi}\)

E. \(\frac{3A}{\pi}\)


Show Spoiler
Attachment:
GFPS2110.jpg
GFPS2110.jpg [ 2.77 KiB | Viewed 3492 times ]
­
ShowHide Answer
Official Answer
B
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 04 Jun 2026
Posts: 111,096
Own Kudos:
818,718
Given Kudos: 106,641
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 111,096
Kudos: 818,718
The circle with center C shown above is tangent to both axes and has 22 Aug 2019, 01:19
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
The circle with center C shown above is tangent to both axes and has an area of A. What is the distance from O to C, in terms of A?

A. A

B. \(\sqrt{\frac{2A}{\pi}}\)

C. \(\sqrt{\frac{3A}{\pi}}\)

D. \(\frac{2A}{\pi}\)

E. \(\frac{3A}{\pi}\)
Show more
Similar question: https://gmatclub.com/forum/the-circle-w ... 44895.html­
User avatar
firas92
User avatar
Current Student
Joined: 16 Jan 2019
Last visit: 02 Dec 2024
Posts: 616
Own Kudos:
1,782
 [4]
Given Kudos: 142
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
Products:
My Reviews
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
Posts: 616
Kudos: 1,782
 [4]
The circle with center C shown above is tangent to both axes and has 22 Aug 2019, 01:55
2
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Draw perpendiculars from C to the x and y axes. We now have a square with side \(r\)(radius of circle) and diagonal \(OC=\sqrt{2}r\)

We know that Area, \(A=\pi r^2\) and so \(r=\sqrt{\frac{A}{\pi}}\)

So \(OC=\sqrt{2}*\sqrt{\frac{A}{\pi}}=\sqrt{\frac{2A}{\pi}}\)

Answer is (B)

Posted from my mobile device
User avatar
Archit3110
User avatar
Major Poster
Joined: 18 Aug 2017
Last visit: 03 Jun 2026
Posts: 8,665
Own Kudos:
5,206
Given Kudos: 243
Status:You learn more from failure than from success.
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1: 545 Q79 V79 DI73
GMAT Focus 2: 645 Q83 V82 DI81
GPA: 4
WE:Marketing (Energy)
GMAT Focus 2: 645 Q83 V82 DI81
Posts: 8,665
Kudos: 5,206
The circle with center C shown above is tangent to both axes and has 22 Aug 2019, 04:42
Kudos
Add Kudos
Bookmarks
Bookmark this Post
let center from x axis be (3,3) so distance o to c ; 3√2
plug in a = 9pi
we get B is correct
IMO B
Bunuel

The circle with center C shown above is tangent to both axes and has an area of A. What is the distance from O to C, in terms of A?

A. A

B. \(\sqrt{\frac{2A}{\pi}}\)

C. \(\sqrt{\frac{3A}{\pi}}\)

D. \(\frac{2A}{\pi}\)

E. \(\frac{3A}{\pi}\)


Show Spoiler
Attachment:
GFPS2110.jpg
Show more
­
User avatar
ScottTargetTestPrep
User avatar
Target Test Prep Representative
Joined: 14 Oct 2015
Last visit: 03 Jun 2026
Posts: 22,403
Own Kudos:
26,621
 [1]
Given Kudos: 302
Status:Founder & CEO
Affiliations: Target Test Prep
Location: United States (CA)
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 22,403
Kudos: 26,621
 [1]
The circle with center C shown above is tangent to both axes and has 24 Aug 2019, 05:28
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

The circle with center C shown above is tangent to both axes and has an area of A. What is the distance from O to C, in terms of A?

A. A

B. \(\sqrt{\frac{2A}{\pi}}\)

C. \(\sqrt{\frac{3A}{\pi}}\)

D. \(\frac{2A}{\pi}\)

E. \(\frac{3A}{\pi}\)





Show Spoiler
Attachment:
GFPS2110.jpg
Show more

Since the area is A, we have:

A = πr^2

A/π = r^2

√(A/π) = r

Note that if we drop a perpendicular from C to the x axis (call it B), then we have created triangle OCB that is a 45-45-90 triangle with legs r and hypotenuse r√2 (which is also the distance from O to C). Recall that the side ratio for a 45-45-90 triangle is x : x : x√2 .

Since the radius is √(A/π), the distance from C to O is √(A/π) * √2 = √(2A/π).

Answer: B
avatar
tempsamarth
avatar
Joined: 28 Aug 2019
Last visit: 15 Jan 2020
Posts: 9
Own Kudos:
5
Given Kudos: 5
Posts: 9
Kudos: 5
The circle with center C shown above is tangent to both axes and has 01 Nov 2019, 01:48
Kudos
Add Kudos
Bookmarks
Bookmark this Post
OC = \(r\sqrt{2}\) (Pythagoras theorem, back to 7th grade)

Area (A) = \(\pi*r^2\) (not sure how to write r squared)

So, \(\sqrt{(A/\pi)} = r\)

OC = \(\sqrt{(A/π)}*\sqrt{2}.\)

Answer is (B)


Please check the approach Bunuel. Thanks :)
Moderator:
Bunuel
Math Expert
111096 posts