Last visit was: 20 Nov 2025, 00:12 It is currently 20 Nov 2025, 00:12
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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 Nov 2025
Posts: 105,408
Own Kudos:
778,417
 [7]
Given Kudos: 99,987
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: 105,408
Kudos: 778,417
 [7]
The circle defined by x^2 + y^2 - 6x - 10y + k = 0 has no common point 05 May 2020, 04:28
Kudos
Add Kudos
7
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

25% (medium)

Question Stats:

76% (02:04) correct 24% (02:14) wrong based on 55 sessions

History

Date
Time
Result
Not Attempted Yet
The circle defined by \(x^2 + y^2 - 6x - 10y + k = 0\) has no common point with any of the axis. If the point (1, 4) does not lie outside the circle, which of the following gives the range of all possible values of k ?


A. \(0 < k < 5\)

B. \(5 \leq k < 10\)

C. \(10 \leq k < 15\)

D. \(15 \leq k \leq 25\)

E. \(25 < k \leq 29\)


Are You Up For the Challenge: 700 Level Questions
ShowHide Answer
Official Answer
E
User avatar
GMATWhizTeam
User avatar
GMATWhiz Representative
Joined: 07 May 2019
Last visit: 14 Oct 2025
Posts: 3,380
Own Kudos:
2,141
 [1]
Given Kudos: 69
Location: India
GMAT 1: 740 Q50 V41
GMAT 2: 760 Q51 V40
Expert
Expert reply
GMAT 2: 760 Q51 V40
Posts: 3,380
Kudos: 2,141
 [1]
The circle defined by x^2 + y^2 - 6x - 10y + k = 0 has no common point Updated on: 05 May 2020, 20:58
Originally posted by GMATWhizTeam on 05 May 2020, 05:17.
Last edited by GMATWhizTeam on 05 May 2020, 20:58, edited 1 time in total.
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
The circle defined by \(x^2 + y^2 - 6x - 10y + k = 0\) has no common point with any of the axis. If the point (1, 4) does not lie outside the circle, which of the following gives the range of all possible values of k ?


A. \(0 < k < 5\)

B. \(5 \leq k < 10\)

C. \(10 \leq k < 15\)

D. \(15 \leq k \leq 25\)

E. \(25 < k \leq 29\)

Are You Up For the Challenge: 700 Level Questions
Show more

Solution



    • Given circle’s equation \(= x^2 + y^2 – 6x – 10y + k = 0\)
      o Or, \(x^2 – 2*3*x + 9 + y^2 – 2*5*y + 25 -9-25 + k = 0\)
      o Or, \((x – 3)^2 + (y – 5)^2 = (\sqrt{34-k})^2\)
    • Thus, the given circle has centre at (3, 5) and has a radius of \(\sqrt{34-k}\)
    • Since, the circle doesn’t touch any of the axis, so radius of the circle must be smaller than the distance between its centre and the nearest axis.
      o This means, \(\sqrt{34-k} < 3\)
         Or, \(34 – k < 9\)
         Or, \(k > 25 ………….(i)\)
    • Also, it is given that the Point (1, 4) doesn’t lie outside the circle. So it must lie either inside the circle or on the circle.
    • So, distance between centre of the circles and point (1, 4) must be less than or equal to the radius.
      o i.e. \((1-3)^2 +(4-5)^2 ≤ 34 – k\)
      o Or, \(4 + 1 ≤ 34 – k\)
      o Or, \(k ≤ 29 ……………(ii)\)
    • Combining inequalities (i) and (ii), we get ,
      o \(25 < k ≤ 29\)
Thus, the correct answer is Option E.
User avatar
lacktutor
User avatar
Joined: 25 Jul 2018
Last visit: 23 Oct 2023
Posts: 659
Own Kudos:
1,396
Given Kudos: 69
Posts: 659
Kudos: 1,396
The circle defined by x^2 + y^2 - 6x - 10y + k = 0 has no common point 05 May 2020, 17:43
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMATWhizTeam
Bunuel
The circle defined by \(x^2 + y^2 - 6x - 10y + k = 0\) has no common point with any of the axis. If the point (1, 4) does not lie outside the circle, which of the following gives the range of all possible values of k ?


A. \(0 < k < 5\)

B. \(5 \leq k < 10\)

C. \(10 \leq k < 15\)

D. \(15 \leq k \leq 25\)

E. \(25 < k \leq 29\)

Are You Up For the Challenge: 700 Level Questions

Solution



    • Given circle’s equation \(= x^2 + y^2 – 6x – 10y + k = 0\)
      o Or, \(x^2 – 2*3*x + 9 + y^2 – 2*5*y + 25 -9-25 + k = 0\)
      o Or, \((x – 3)^2 + (y – 5)^2 = (\sqrt{34-k})^2\)
    • Thus, the given circle has centre at (3, 5) and has a radius of \(\sqrt{34-k}\)
    • Since, the circle doesn’t touch any of the axis, so radius of the circle must be smaller than the distance between its centre and the nearest axis.
      o This means, \(\sqrt{34-k} < 3\)
         Or, \(34 – k < 9\)
         Or, \(k > 25 ………….(i)\)
    • Also, it is given that the Point (1, 4) doesn’t lie outside the circle. So it must lie either inside the circle or on the circle.
    • So, distance between centre of the circles and point (1, 4) must be less than or equal to the radius.
      o i.e. \((1-3)^2 +(4-5)^2 ≤ 34 – k\)
      o Or, \(4 + 1 ≤ 34 – k\)
      o Or, \(k ≤ 29 ……………(ii)\)
    • Combining inequalities (i) and (ii), we get ,
      o \(25 < k ≤ 29\)
Thus, the correct answer is Option D.
Show more

hi, GMATWhizTeam
I really liked your explanation.
But I think there is a typo in your answer option.
Shouldn't it be E ?
User avatar
GMATWhizTeam
User avatar
GMATWhiz Representative
Joined: 07 May 2019
Last visit: 14 Oct 2025
Posts: 3,380
Own Kudos:
2,141
Given Kudos: 69
Location: India
GMAT 1: 740 Q50 V41
GMAT 2: 760 Q51 V40
Expert
Expert reply
GMAT 2: 760 Q51 V40
Posts: 3,380
Kudos: 2,141
The circle defined by x^2 + y^2 - 6x - 10y + k = 0 has no common point 05 May 2020, 21:02
Kudos
Add Kudos
Bookmarks
Bookmark this Post
lacktutor

I really liked your explanation.
But I think there is a typo in your answer option.
Shouldn't it be E ?
Show more


Thanks for pointing it out. I have edited the answer option. :)
Moderators:
Bunuel
Math Expert
105408 posts
Krunaal
Tuck School Moderator
805 posts