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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

29 Dec 2012, 05:28

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

88% (02:20) correct
12% (01:16) wrong based on 673 sessions

HideShow timer Statistics

Attachment:

Circle.png [ 3.73 KiB | Viewed 12579 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) \(\frac{k}{3}\)

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

Show Tags

29 Dec 2012, 05:32

3

This post received KUDOS

Expert's post

4

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 11053 times ]

Since OC=k, then \(r^2+r^2=k^2\) --> \(r=\frac{k}{\sqrt{2}}\).

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

Show Tags

04 Nov 2013, 17:30

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.

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

Show Tags

31 Dec 2014, 07:14

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. _________________

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

Show Tags

31 Dec 2014, 12:09

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.

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

Show Tags

09 Feb 2016, 13:47

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?

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

Show Tags

10 Feb 2016, 05:20

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: \(k^2 = r^2 + r^2\) ----> \(k^2 = 2r^2\) ----> \(\frac{k^2}{2} = r^2\) ----> \(\sqrt{\frac{k^2}{2}} = \sqrt{r^2}\) ----> \(\frac{k}{\sqrt{2}} = r\)

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

Show Tags

10 Feb 2016, 10:38

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).

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

Show Tags

30 Apr 2016, 19:58

Walkabout wrote:

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.)

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

Show Tags

02 May 2016, 09:20

Walkabout wrote:

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.

The answer is B. _________________

Jeffrey Miller Scott Woodbury-Stewart Founder and CEO

gmatclubot

Re: The circle with center C shown above is tangent to both axes
[#permalink]
02 May 2016, 09:20

http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...