Find all School-related info fast with the new School-Specific MBA Forum

It is currently 13 Feb 2016, 20:01
GMAT Club Tests

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

The circle with center C shown above is tangent to both axes

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 02 Dec 2012
Posts: 178
Followers: 3

Kudos [?]: 1379 [0], given: 0

The circle with center C shown above is tangent to both axes [#permalink] New post 29 Dec 2012, 04:28
5
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

88% (02:22) correct 12% (01:20) wrong based on 580 sessions
Attachment:
Circle.png
Circle.png [ 3.73 KiB | Viewed 11036 times ]
The circle with center C shown above is tangent to both axes. If the distance from O to C is equal to k, what is the radius of the circle, in terms of k ?

(A) k
(B) \(\frac{k}{\sqrt{2}}\)
(C) \(\frac{k}{\sqrt{3}}\)
(D) \(\frac{k}{2}\)
(E) \(\frac{k}{3}\)
[Reveal] Spoiler: OA
Expert Post
3 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 31304
Followers: 5365

Kudos [?]: 62566 [3] , given: 9457

Re: The circle with center C shown above is tangent to both axes [#permalink] New post 29 Dec 2012, 04:32
3
This post received
KUDOS
Expert's post
3
This post was
BOOKMARKED
Image
The circle with center C shown above is tangent to both axes. If the distance from O to C is equal to k, what is the radius of the circle, in terms of k ?

(A) k
(B) \(\frac{k}{\sqrt{2}}\)
(C) \(\frac{k}{\sqrt{3}}\)
(D) \(\frac{k}{2}\)
(E) \(\frac{k}{3}\)

Look at the diagram below:
Attachment:
Circle2.png
Circle2.png [ 5.4 KiB | Viewed 9676 times ]
Since OC=k, then \(r^2+r^2=k^2\) --> \(r=\frac{k}{\sqrt{2}}\).

Answer: B.
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 12 Feb 2013
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 8

Re: The circle with center C shown above is tangent to both axes [#permalink] New post 20 Jun 2013, 17:04
How did we conclude that the base of the triangle is also r ?
6 KUDOS received
Intern
Intern
User avatar
Status: Attempting once more
Joined: 20 Jun 2013
Posts: 1
Location: India
GMAT 1: 700 Q50 V34
GPA: 3.15
Followers: 1

Kudos [?]: 11 [6] , given: 0

Re: The circle with center C shown above is tangent to both axes [#permalink] New post 20 Jun 2013, 17:26
6
This post received
KUDOS
kkbimalnair wrote:
How did we conclude that the base of the triangle is also r ?



Attachment:
Untitled.jpg
Untitled.jpg [ 6.15 KiB | Viewed 9044 times ]


AC is the radius of the circle and Y-axis is tangent to the circle.Which implies that AC is perpendicular to the Y-axis.

Similarly BC is perpendicular to the X-axis.

Angle AOB is right angle. Which implies angle ACB is also right angled.

Now the quadrilateral OACB must be either a rectangle or a Square.

In either case OB must be equal to AC(as both in a rectangle and in a square opposite sides are equal) . So OB = R
_________________

GMAT is not an exam..Its a life style

Current Student
avatar
Joined: 26 Sep 2013
Posts: 221
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41
GMAT 2: 730 Q49 V41
Followers: 3

Kudos [?]: 94 [0], given: 40

Re: The circle with center C shown above is tangent to both axes [#permalink] New post 04 Nov 2013, 16:30
I actually solved this one with a bit of logic, or at least eliminated several answer choices.

A) is obviously out since k includes that area outside of the circle by the origin, so the radius must be shorter. Dividing it by 2 or 3 seemed overkill. dividing k by the sqrt of 3 implied some sort of 30/60/90 triangle which didn't make much sense. so I figured it was B, then made the triangle real quick, saw it was a right triangle and as soon as I wrote out the pythag equation I saw that B was definitely correct.
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 8240
Followers: 419

Kudos [?]: 111 [0], given: 0

Top 10 in overall
Re: The circle with center C shown above is tangent to both axes [#permalink] New post 31 Dec 2014, 06:14
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Expert Post
EMPOWERgmat Instructor
User avatar
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 5623
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Followers: 227

Kudos [?]: 1603 [0], given: 154

Re: The circle with center C shown above is tangent to both axes [#permalink] New post 31 Dec 2014, 11:09
Expert's post
Hi All,

This question can be solved by TESTing VALUES.

First though, there's a hidden pattern worth noting: Since the circle is tangent to both the x-axis and the y-axis, we can draw a SQUARE into the picture (using the Origin as one corner and "C" as the opposite corner). The length of a side of this square will EQUAL the RADIUS of the circle. From here, we can make the dimensions of the square anything that we want.

Let's TEST...

Radius = 2

The length from the Origin to the center C cuts through the square and forms two 45/45/90 right triangles. Thus...

K = That length = 2\sqrt{2}

Now we just have to plug THAT value into the answers to find the one that equals 2.

The only answer that matches is
[Reveal] Spoiler:
B


GMAT assassins aren't born, they're made,
Rich
_________________

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests

60-point improvement guarantee
www.empowergmat.com/

Intern
Intern
avatar
Joined: 13 Sep 2015
Posts: 22
Followers: 0

Kudos [?]: 4 [0], given: 239

The circle with center C shown above is tangent to both axes [#permalink] New post 04 Dec 2015, 07:10
alternate method, similar to using the right angle hypotenuese formula:

notice that the figure also forms a cube with r (radius) as each side, and diagonal is k or CO.
Using diagonal of square formula
r√2 = k
r = k/√2

(diagonal of a square is x√2)
Intern
Intern
avatar
Joined: 13 Jun 2011
Posts: 12
Followers: 0

Kudos [?]: 0 [0], given: 13

Re: The circle with center C shown above is tangent to both axes [#permalink] New post 07 Dec 2015, 05:57
Mantis wrote:
kkbimalnair wrote:
How did we conclude that the base of the triangle is also r ?



Attachment:
Untitled.jpg


AC is the radius of the circle and Y-axis is tangent to the circle.Which implies that AC is perpendicular to the Y-axis.

Similarly BC is perpendicular to the X-axis.

Angle AOB is right angle. Which implies angle ACB is also right angled.

Now the quadrilateral OACB must be either a rectangle or a Square.

In either case OB must be equal to AC(as both in a rectangle and in a square opposite sides are equal) . So OB = R



Hi Can you please let me know how you made AOB as right angle?
Intern
Intern
avatar
Joined: 03 Jan 2015
Posts: 41
Followers: 0

Kudos [?]: 5 [0], given: 117

GMAT ToolKit User
Re: The circle with center C shown above is tangent to both axes [#permalink] New post 09 Feb 2016, 12:47
Could someone explain me why I cannot take the square root of this entire expression:
\(\sqrt{k^2 = r^2 + r^2}\) ---> \(k = r + r.\)
Why is this incorrect?
Expert Post
1 KUDOS received
EMPOWERgmat Instructor
User avatar
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 5623
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Followers: 227

Kudos [?]: 1603 [1] , given: 154

Re: The circle with center C shown above is tangent to both axes [#permalink] New post 09 Feb 2016, 18:59
1
This post received
KUDOS
Expert's post
Hi saiesta,

You have to combine 'like' terms before you take the square-root of both sides.

Here's a simple example that proves WHY:

\(\sqrt{4}\) = 2

\(\sqrt{(2+2)}\) does NOT = \(\sqrt{2}\) + \(\sqrt{2}\) though

\(\sqrt{2}\) = about 1.4

2 does NOT equal 1.4 + 1.4

Knowing THAT, how would you write your equation now?

GMAT assassins aren't born, they're made,
Rich
_________________

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests

60-point improvement guarantee
www.empowergmat.com/

1 KUDOS received
Intern
Intern
avatar
Joined: 03 Jan 2015
Posts: 41
Followers: 0

Kudos [?]: 5 [1] , given: 117

GMAT ToolKit User
Re: The circle with center C shown above is tangent to both axes [#permalink] New post 10 Feb 2016, 04:20
1
This post received
KUDOS
EMPOWERgmatRichC wrote:
Hi saiesta,

You have to combine 'like' terms before you take the square-root of both sides.

Here's a simple example that proves WHY:

\(\sqrt{4}\) = 2

\(\sqrt{(2+2)}\) does NOT = \(\sqrt{2}\) + \(\sqrt{2}\) though

\(\sqrt{2}\) = about 1.4

2 does NOT equal 1.4 + 1.4

Knowing THAT, how would you write your equation now?

GMAT assassins aren't born, they're made,
Rich

Thank you, I have to remember myself to combine like terms before I take the square root of both sides. I believe this is particularly the case when adding or subtracting (square) roots.
I would write the equation as follows: \(k^2 = r^2 + r^2\) ----> \(k^2 = 2r^2\) ----> \(\frac{k^2}{2} = r^2\) ----> \(\sqrt{\frac{k^2}{2}} = \sqrt{r^2}\) ----> \(\frac{k}{\sqrt{2}} = r\)
Expert Post
1 KUDOS received
EMPOWERgmat Instructor
User avatar
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 5623
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Followers: 227

Kudos [?]: 1603 [1] , given: 154

Re: The circle with center C shown above is tangent to both axes [#permalink] New post 10 Feb 2016, 09:38
1
This post received
KUDOS
Expert's post
Hi saiesta,

Nicely done. As a minor aside, since this is a Geometry question, you don't have to worry about any of the 'measures' ending up negative. However, if the same concepts were in an Algebraic prompt, then you'd have to keep in mind that both K and R could be negative (so square-rooting both sides might end up eliminating possible answers; in a DS question, that could lead to a wrong answer).

GMAT assassins aren't born, they're made,
Rich
_________________

Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests

60-point improvement guarantee
www.empowergmat.com/

Re: The circle with center C shown above is tangent to both axes   [#permalink] 10 Feb 2016, 09:38
    Similar topics Author Replies Last post
Similar
Topics:
Experts publish their posts in the topic If the center of a circle lies on a smaller circle, as shown above, wh reto 3 27 Jun 2015, 04:42
If P is the center of the circle shown above, and BAC=30º, and the are reto 3 09 May 2015, 01:52
2 If P is the center of the circle shown above, and BAC=30º, and the are reto 1 09 May 2015, 00:59
Experts publish their posts in the topic If points A and C both lie on the circle with center B and jessello 4 09 Dec 2012, 00:46
24 Experts publish their posts in the topic If A is the center of the circle shown above (see attachment TheRob 18 07 Sep 2009, 06:19
Display posts from previous: Sort by

The circle with center C shown above is tangent to both axes

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.