Last visit was: 19 Nov 2025, 07:48 It is currently 19 Nov 2025, 07:48
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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
   1   2 
Sub 505 Level|   Geometry|                        
User avatar
dave13
User avatar
Joined: 09 Mar 2016
Last visit: 12 Aug 2025
Posts: 1,108
Own Kudos:
1,113
Given Kudos: 3,851
Posts: 1,108
Kudos: 1,113
The circle with center C shown above is tangent to both axes 02 Mar 2018, 11:18
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

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

Answer: B.
Show more


Dear Bunuel, hello ! :-)

i know you are tired of my questions :) but i am sure you still can understand me :)

can you please explan why do you use this formula? \(r^2+r^2=k^2\)

from this post https://gmatclub.com/forum/math-coordin ... 87652.html

i know that the if the circle is centered at the origin (0, 0), then the equation simplifies to: \(x^2+y^2 = r^2\) and in other cases

we use this formula In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:

\((x−a)^2+(y−b)^2=r\)
User avatar
dave13
User avatar
Joined: 09 Mar 2016
Last visit: 12 Aug 2025
Posts: 1,108
Own Kudos:
1,113
Given Kudos: 3,851
Posts: 1,108
Kudos: 1,113
The circle with center C shown above is tangent to both axes 02 Mar 2018, 12:13
Kudos
Add Kudos
Bookmarks
Bookmark this Post
EMPOWERgmatRichC
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
Show more

Hello EMPOWERgmatRichC, :)

what do you mean by combining 'like' terms before you take the square-root of both sides :? dont understand ... :-)
User avatar
dave13
User avatar
Joined: 09 Mar 2016
Last visit: 12 Aug 2025
Posts: 1,108
Own Kudos:
1,113
Given Kudos: 3,851
Posts: 1,108
Kudos: 1,113
The circle with center C shown above is tangent to both axes 03 Mar 2018, 13:28
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel

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

Answer: B.
Show more

Bunuel shouldnt we rationalise denominator :) \(r=\frac{k}{\sqrt{2}}\).

for instance if we have \(\frac{\sqrt{6}}{\sqrt{2}}\) = \(\frac{\sqrt{6}}{\sqrt{2}}\) *\(\frac{\sqrt{2}}{\sqrt{2}}\) = \(\frac{\sqrt{12}}{\sqrt{4}}\)
so we get \(\frac{\sqrt{12}}2\)

As you see in the denominator 2 is without radical sign, so why you didn't write it so ?

Another question sqrt of 2 is 1.4 you could write it so in the denominator too ? in which case you could write 1.4 ? can you please explain ? :) pleaseee:)
User avatar
Kimberly77
User avatar
Joined: 16 Nov 2021
Last visit: 07 Sep 2024
Posts: 435
Own Kudos:
45
Given Kudos: 5,898
Location: United Kingdom
GMAT 1: 450 Q42 V34
Products:
My Reviews
GMAT 1: 450 Q42 V34
Posts: 435
Kudos: 45
The circle with center C shown above is tangent to both axes 20 Nov 2022, 12:47
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Walkabout
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) \(\frac{k}{3}\)
Show more

Hi Brent BrentGMATPrepNow, somehow I ended up connecting line from C to X. Would it still works this way? Thanks Brent
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,739
Own Kudos:
35,341
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,739
Kudos: 35,341
The circle with center C shown above is tangent to both axes 20 Nov 2022, 13:32
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Kimberly77


Hi Brent BrentGMATPrepNow, somehow I ended up connecting line from C to X. Would it still works this way? Thanks Brent
Show more

I'm not sure what you mean.
"x" isn't a point on the x-axis; the "x" just lets us know that the horizontal line is the x axis.

If you show me your calculations, I'd be happy to take a look.
User avatar
Kimberly77
User avatar
Joined: 16 Nov 2021
Last visit: 07 Sep 2024
Posts: 435
Own Kudos:
45
Given Kudos: 5,898
Location: United Kingdom
GMAT 1: 450 Q42 V34
Products:
My Reviews
GMAT 1: 450 Q42 V34
Posts: 435
Kudos: 45
The circle with center C shown above is tangent to both axes 21 Nov 2022, 11:36
Kudos
Add Kudos
Bookmarks
Bookmark this Post
BrentGMATPrepNow
Kimberly77


Hi Brent BrentGMATPrepNow, somehow I ended up connecting line from C to X. Would it still works this way? Thanks Brent

I'm not sure what you mean.
"x" isn't a point on the x-axis; the "x" just lets us know that the horizontal line is the x axis.

If you show me your calculations, I'd be happy to take a look.
Show more

Thanks BrentGMATPrepNow. I see what you mean now that I've mistaken X as a point to draw the line to.
User avatar
ArnauG
User avatar
Joined: 23 Dec 2022
Last visit: 14 Oct 2023
Posts: 298
Own Kudos:
42
Given Kudos: 199
Posts: 298
Kudos: 42
The circle with center C shown above is tangent to both axes 03 Jun 2023, 05:10
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Given that the distance from the origin O to center C is equal to k, we need to find the radius of the circle in terms of k.

Since the circle is tangent to both axes, the radius will be the length of the line segment from the center C to either of the points of tangency on the axes.

Let's consider the point of tangency on the x-axis as A. Since the circle is tangent to the x-axis, the radius and the y-coordinate of point C will be equal.

Now, we have a right triangle OCA, with OC as the hypotenuse (length k) and AC as one of the legs (radius of the circle).

Applying the Pythagorean theorem, we have:

k^2 = AC^2 + OA^2

Since AC is the radius of the circle and OA is the distance from the origin to the point of tangency on the x-axis, which is equal to the radius of the circle as well, we can rewrite the equation as:

k^2 = AC^2 + AC^2
2AC^2 = k^2
AC^2 = k^2 / 2

Taking the square root of both sides, we have:

AC = sqrt(k^2 / 2) = k / sqrt(2)

Therefore, the radius of the circle is equal to AC, which is k divided by the square root of 2.

Hence, the correct answer is (B) k/√2.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,588
Own Kudos:
1,079
Posts: 38,588
Kudos: 1,079
The circle with center C shown above is tangent to both axes 12 Jun 2025, 00:57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
   1   2 
Moderators:
Bunuel
Math Expert
105389 posts
Krunaal
Tuck School Moderator
805 posts