It is currently 16 Jan 2018, 03:52

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 02 Dec 2012
Posts: 178

Kudos [?]: 3762 [0], given: 0

The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

29 Dec 2012, 04:28
17
This post was
BOOKMARKED
00:00

Difficulty:

5% (low)

Question Stats:

87% (01:14) correct 13% (01:15) wrong based on 1463 sessions

### HideShow timer Statistics

Attachment:

Circle.png [ 3.73 KiB | Viewed 25935 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

Kudos [?]: 3762 [0], given: 0

Math Expert
Joined: 02 Sep 2009
Posts: 43292

Kudos [?]: 139155 [3], given: 12777

Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

29 Dec 2012, 04:32
3
KUDOS
Expert's post
8
This post was
BOOKMARKED

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 [ 5.4 KiB | Viewed 21885 times ]
Since OC=k, then $$r^2+r^2=k^2$$ --> $$r=\frac{k}{\sqrt{2}}$$.

_________________

Kudos [?]: 139155 [3], given: 12777

Intern
Joined: 12 Feb 2013
Posts: 1

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

Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

20 Jun 2013, 17:04
How did we conclude that the base of the triangle is also r ?

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

Intern
Status: Attempting once more
Joined: 20 Jun 2013
Posts: 1

Kudos [?]: 13 [7], given: 0

Location: India
GMAT 1: 700 Q50 V34
GPA: 3.15
Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

20 Jun 2013, 17:26
7
KUDOS
1
This post was
BOOKMARKED
kkbimalnair wrote:
How did we conclude that the base of the triangle is also r ?

Attachment:

Untitled.jpg [ 6.15 KiB | Viewed 21233 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

Kudos [?]: 13 [7], given: 0

Manager
Joined: 26 Sep 2013
Posts: 217

Kudos [?]: 190 [0], given: 40

Concentration: Finance, Economics
GMAT 1: 670 Q39 V41
GMAT 2: 730 Q49 V41
Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

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.

Kudos [?]: 190 [0], given: 40

Non-Human User
Joined: 09 Sep 2013
Posts: 14282

Kudos [?]: 291 [0], given: 0

Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

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.
_________________

Kudos [?]: 291 [0], given: 0

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 10673

Kudos [?]: 3771 [0], given: 173

Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

31 Dec 2014, 11:09
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...

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
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Kudos [?]: 3771 [0], given: 173 Intern Joined: 13 Sep 2015 Posts: 19 Kudos [?]: 10 [0], given: 239 The circle with center C shown above is tangent to both axes [#permalink] ### Show Tags 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) Kudos [?]: 10 [0], given: 239 Intern Joined: 13 Jun 2011 Posts: 24 Kudos [?]: [0], given: 27 Re: The circle with center C shown above is tangent to both axes [#permalink] ### Show Tags 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? Kudos [?]: [0], given: 27 Manager Joined: 03 Jan 2015 Posts: 86 Kudos [?]: 69 [0], given: 146 Re: The circle with center C shown above is tangent to both axes [#permalink] ### Show Tags 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? Kudos [?]: 69 [0], given: 146 EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 10673 Kudos [?]: 3771 [1], given: 173 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: The circle with center C shown above is tangent to both axes [#permalink] ### Show Tags 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 _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Kudos [?]: 3771 [1], given: 173

Manager
Joined: 03 Jan 2015
Posts: 86

Kudos [?]: 69 [1], given: 146

Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

10 Feb 2016, 04:20
1
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: $$k^2 = r^2 + r^2$$ ----> $$k^2 = 2r^2$$ ----> $$\frac{k^2}{2} = r^2$$ ----> $$\sqrt{\frac{k^2}{2}} = \sqrt{r^2}$$ ----> $$\frac{k}{\sqrt{2}} = r$$

Kudos [?]: 69 [1], given: 146

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 10673

Kudos [?]: 3771 [1], given: 173

Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

10 Feb 2016, 09:38
1
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
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Kudos [?]: 3771 [1], given: 173

Intern
Joined: 10 Aug 2015
Posts: 31

Kudos [?]: 7 [0], given: 231

Location: India
GMAT 1: 700 Q48 V38
GPA: 3.5
WE: Consulting (Computer Software)
Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

30 Apr 2016, 18:58
Attachment:
Circle.png
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) $$\frac{k}{3}$$

If we draw radius to two point of tangency(we will have two 90 degrees there and other two angles will also be 90 degrees. find out why.) then we will draw a square with 4 sides equal(find out why ). Now we have diagonal of square = k . then side = k/sqrt2.(why?= 45-45-90 right angle triangle.)

Kudos [?]: 7 [0], given: 231

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 1997

Kudos [?]: 1080 [2], given: 4

Location: United States (CA)
Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

02 May 2016, 08:20
2
KUDOS
Expert's post
Attachment:
Circle.png
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) $$\frac{k}{3}$$

We begin by creating a right triangle with the x-axis, the radius of circle C to the x-axis and the line segment from center of circle C to the origin. We see that we have created an isosceles right triangle, also known as a 45-45-90 degree right triangle. We know this because each leg of the right triangle is equal to a radius of the circle. We can label all this in our diagram.

We know the side-hypotenuse ratios in a 45-45-90 degree right triangle are:

x:x:x√2, where x represents the leg of the triangle and x√2 is the hypotenuse.

We can use this to determine the leg of the triangle.

Since x√2 equals the hypotenuse of the triangle we can say:

x√2 = k

x = k/√2

Since x also represents the radius of the circle, k/√2 is equal to the radius.

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Kudos [?]: 1080 [2], given: 4

Intern
Joined: 15 Sep 2016
Posts: 5

Kudos [?]: 5 [0], given: 15

Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

04 Oct 2016, 23:26
sin45=r/k
1/√2 = r/k
r=k/√2

Kudos [?]: 5 [0], given: 15

Intern
Joined: 16 May 2017
Posts: 21

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

Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

18 May 2017, 08:09
Based on Pythagorean theorem,
r = k/$$\sqrt{2}$$

Attachments

Screen Shot 2017-05-18 at 11.56.46 AM.png [ 41.98 KiB | Viewed 5335 times ]

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

Senior Manager
Status: Professional GMAT Tutor
Affiliations: AB, cum laude, Harvard University (Class of '02)
Joined: 10 Jul 2015
Posts: 477

Kudos [?]: 580 [0], given: 63

Location: United States (CA)
Age: 38
GMAT 1: 770 Q47 V48
GMAT 2: 730 Q44 V47
GMAT 3: 750 Q50 V42
GRE 1: 337 Q168 V169
WE: Education (Education)
Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

30 May 2017, 15:26
Top Contributor
Attached is a visual that should help.
Attachments

Screen Shot 2017-05-30 at 4.07.38 PM.png [ 88.7 KiB | Viewed 5066 times ]

_________________

Harvard grad and 770 GMAT scorer, offering high-quality private GMAT tutoring, both in-person and online via Skype, since 2002.

GMAT Action Plan - McElroy Tutoring

Kudos [?]: 580 [0], given: 63

Director
Joined: 02 Sep 2016
Posts: 784

Kudos [?]: 24 [0], given: 275

Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

04 Sep 2017, 10:37
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

Bunuel, abhimahna, and experts

Is my understanding of this question correct ?

As two tangents drawn from the same point are equal, then OA=OB.

Radius AC and CB are perpendicular to the two tangents and make 90 degree angle.

Angle AOB is also 90 degrees because y axis is perpendicular to x axis.

So angle ACB is also 90 degree.

If ACBO is a rectangle, then opposite sides are equal. Also as two tangents are equal and both the radius are equal, its a square.

Triangle AOB is a 45-45-90 triangle and hence k= root 2 and AO=OB=BC=AC= 1

Applying the theorem:

r^2+r^2= k^2
2r^2= k^2

r= k/root 2

Kudos [?]: 24 [0], given: 275

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4669

Kudos [?]: 3301 [0], given: 0

GPA: 3.82
Re: The circle with center C shown above is tangent to both axes [#permalink]

### Show Tags

05 Sep 2017, 17:21
Attachment:
Circle.png
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) $$\frac{k}{3}$$

We have a right triangle with sides r, r, k. It is an isosceles.

$$k^2 = r^2 + r^2$$
$$k^2 = 2 * r^2$$
$$k = \sqrt{2}*r$$
$$r = k / \sqrt{2}$$

Ans: B
_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
Find a 10% off coupon code for GMAT Club members.
“Receive 5 Math Questions & Solutions Daily”
Unlimited Access to over 120 free video lessons - try it yourself

Kudos [?]: 3301 [0], given: 0

Re: The circle with center C shown above is tangent to both axes   [#permalink] 05 Sep 2017, 17:21

Go to page    1   2    Next  [ 21 posts ]

Display posts from previous: Sort by