BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Points on circle

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 53
Joined: Sat Jul 17, 2010 3:56 am
Thanked: 1 times

by mourinhogmat1 » Mon Jan 09, 2012 8:05 pm
If centre is (0,0) and radius is 5, then for x and y we need to test from -5 to +5 including 0.
But if we just try for x from 0 to 5. We know that potential combos are (5,0); (3,4).
For each number if we have positive and negative except of course 0 we have, 8 possibilities for (3,4) combination i.e. (3,4); (-3,4)...and so on including the reversal when x and y are interchanged.
For (5,0) we have 4 combinations i.e. (5,0); (-5,0) and so on including the reversal when x and y are interchanged.

So, total is 8+4 = 12 possibilities.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 626
Joined: Fri Dec 23, 2011 2:50 amFollowed by: 10 members
Location: Ahmedabad
Thanked: 31 times

by ronnie1985 » Tue Jan 10, 2012 12:40 am
the solution is the no of solutions for the equation x^2+y^2 = 25.
The integral solutions are (0,+/-5) (+/-5,0) (+/-4,+/-3), (+/-3,+/-4). totaling 12 points.
Follow your passion, Success as perceived by others shall follow you
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 626
Joined: Fri Dec 23, 2011 2:50 amFollowed by: 10 members
Location: Ahmedabad
Thanked: 31 times

by ronnie1985 » Fri Mar 30, 2012 9:36 am
(3,4,5) is a pythagorean triplet. Hence 4 point with all positive negative combination of (3,4) and (4,3) are also on the circle.
12 points are integral.
Follow your passion, Success as perceived by others shall follow you
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 183
Joined: Wed Sep 21, 2011 7:06 amFollowed by: 2 members
Location: Washington, DC
Thanked: 6 times
GMAT Score:500

by Rastis » Sat Apr 07, 2012 4:10 am
Can someone tell me why the Pythagorean Theorem was used and how they came up with the points?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 72
Joined: Mon Apr 11, 2011 6:45 pm
Thanked: 2 times

by shubhamkumar » Sat Apr 07, 2012 4:26 am
Rastis wrote:Can someone tell me why the Pythagorean Theorem was used and how they came up with the points?
Well there are 4 possibilities when the radius will lie on the axis(0,-5),(0,5),(5,0)(-5,0).
Now we need to find other points in the co-ordinate planes which have integer values.
You can remember the equation of the circle as x^2+y^2=r^2.Where x and y are the co-ordinates of the point,the circle will pass through and r is the radius,or the distance between the origin and the point(x,y).If you look closely this equation is also the pythagorous equation.
Another way to look at it is suppose you drop a perpendicular from point (x,y) on the x axis to form a triangle with the origin,the base of the perpendicular(0,x) and point x,y.This triangle is a right angled triangle.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 183
Joined: Wed Sep 21, 2011 7:06 amFollowed by: 2 members
Location: Washington, DC
Thanked: 6 times
GMAT Score:500

by Rastis » Sat Apr 07, 2012 6:31 am
Thank you for that clarification, shubhamkumar. I didn't know the equation of a circle so that was one issue. However, I still do not know how the other points were found that add up to 12.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 626
Joined: Fri Dec 23, 2011 2:50 amFollowed by: 10 members
Location: Ahmedabad
Thanked: 31 times

by ronnie1985 » Sat Apr 07, 2012 9:01 am
(3,4,5) are pythagorean triplets
Hence the integer points are (+/-3,+/-4), (+/-4,+/-3) (+/-5,0) and (0,+/-5) total 12 points
Follow your passion, Success as perceived by others shall follow you
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 183
Joined: Wed Sep 21, 2011 7:06 amFollowed by: 2 members
Location: Washington, DC
Thanked: 6 times
GMAT Score:500

by Rastis » Sat Apr 07, 2012 4:49 pm
but arent' there an infinite number of points on a circle? I am confused as to why those points are the only ones that are integers.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 9
Joined: Tue Jul 19, 2011 6:35 amFollowed by: 1 member

by keepintouch_7 » Mon Apr 30, 2012 8:49 am
the center of the circle is origin(0,0).
Any point on the given circle satisfies the equation x^2+y^2 = 5^2(5 being the radius).So all the infinite points on the circle will satisfy this equation.
Now you have to identify the points which will be integer.Start from X=0 so this will give 0^2+y^2=25
which gives Y = +5 and -5 so two points are 0,5 and 0,-5.
Again put x=1 then this will not give integer value for y.
So test X=3 and X=-3 which will give Y=+4 and Y=-4
like wise test for X=+4 and X=-4 and x=+5 and X=-5

So you will get a total of 12 points only having both x and y as integers.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 37
Joined: Mon Oct 10, 2011 2:27 am
Thanked: 2 times

by ritzzzr » Tue Feb 19, 2013 9:49 pm
The points will be the solution for the following eq

x^2+y^2=25
0,+-5,+-4,+-3

we have 12 points
sp 12 is the ans
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 171
Joined: Tue Jan 08, 2013 7:24 am
Thanked: 1 times

by rajeshsinghgmat » Sat Feb 23, 2013 1:20 am
C) 12

x^2+y^2=25
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 234
Joined: Tue Jul 16, 2013 9:00 am
Location: West Virginia
Thanked: 9 times

by Java_85 » Tue Sep 17, 2013 6:48 am
IMO also C.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Junior | Next Rank: 30 Posts
Posts: 16
Joined: Tue Nov 12, 2013 11:20 amFollowed by: 3 members
Location: Appleton, WI
Thanked: 1 times

by marshelle.slayton » Thu Nov 06, 2014 3:51 am
Rastis wrote:but arent' there an infinite number of points on a circle? I am confused as to why those points are the only ones that are integers.
Integers must be whole numbers and in this case there are only 12 options. Think of just one section of this circle and it makes the problem a bit easier. If you think of just one quarter section then multiply the answer by 4 it is easier.

You know the radius is 5 so at the y axis you have (0,5) and the x axis you have (5,0). If triangles are comfortable for you, think of how many different right (90 degree) triangles you can make with integer base and heights with a hypotenuse length of 5.

The common 3-4-5 triangle is the only option. It can be flipped so that the 3 refers to the x or y axis giving us two additional options for the 1/4 section.

Take all of the option so far (4) and take away one of the options that will be replicated (I took away the y-axis (0,5) and multiply the 3 by 4 sections of the circle for 12 total.

That is how I thought of it! Hope it helps!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1100
Joined: Sat May 10, 2014 11:34 pmFollowed by: 24 members
Location: New Delhi, India
Thanked: 205 times

by GMATinsight » Thu Nov 06, 2014 10:15 am
Here is another Question on similar lines for you to practice...

Answer: Option D
Attachments
Untitled.jpg
"GMATinsight"Bhoopendra Singh & Sushma Jha
Most Comprehensive and Affordable Video Course 2000+ CONCEPT Videos and Video Solutions
Whatsapp/Mobile: +91-9999687183 l [email protected]
Contact for One-on-One FREE ONLINE DEMO Class Call/e-mail
Most Efficient and affordable One-On-One Private tutoring fee - US$40-50 per hour
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 518
Joined: Tue May 12, 2015 8:25 pm
Thanked: 10 times

by nikhilgmat31 » Tue Jul 07, 2015 2:10 am
Saying pythagorus theorem is wrong way.

we know the distance of x,y to 0,0 is 5
so
x^2+ y^2 = 5^2

put x as 0,2,3,5 & get y as 5,3,2,0

total of 12 points.

C
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply