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Points on circle

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by diebeatsthegmat » Thu Nov 11, 2010 6:28 pm
vaibhavjha wrote:If the radius of a circle that centers at origin is 5. How many points on the circle have integer coordinates?

a.4
b.8
c.12
d.16
e.20
i supposed that the answer is E because
we can draw a circle with center O and the center will go thru 4 point (5,5), (5,-5), (-5,5), (-5.-=5)
or we can draw it with the centre 5 or -5 from both coordinates x and y. this centre will go thru 8 points
or we can draw it with the centre 5 which coordinates x and y are tangent lines
we also have 8 points
draw coordinate, you will see
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by Rahul@gurome » Thu Nov 11, 2010 7:04 pm
Let any point on the circle be denoted by (x,y).
Since centre of the circle is origin (0, 0) and radius is 5, we get that x^2 + y^2 = 5^2 = 25.
So the possible values of (x, y) are (3, 4), (3, -4), (-3, 4), (-3, -4), (4, 3), (4, -3), (-4, 3), (-4, -3), (5, 0), (-5, 0), (0, 5), (0, -5) where x and y can be integers only.
There are in total 12 points.

The correct answer is c.
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by goyalsau » Thu Nov 11, 2010 8:37 pm
vaibhavjha wrote:If the radius of a circle that centers at origin is 5. How many points on the circle have integer coordinates?

a.4
b.8
c.12
d.16
e.20
there will be 12 coordinates in all,

we know the pythagoras theorem ( 3 ) ^ 2 + ( 4 ) ^ 2 = ( 5 ) ^ 2


in Ist quadrant there will be 3,4 or 4,3
so in all 4 quadrant there will be 8

and 4 at the four lines..
Saurabh Goyal
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by vishubn » Thu Nov 11, 2010 9:38 pm
i really dint understand this , cvan u please explain more?
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by maoriba » Fri Nov 12, 2010 4:19 am
hey...the issue is quite easy.
equation if circle is (x-xC)^2 + (y-yC)^2 = 5^2, right?
so, considering the centre is in 0, 0, xC and Yc are equal to 0, the equation turns into x^2 + y^2 = 25.

Which can be the values of x & y to satisfy this equation?

if x= 0, the equation egts transformed into y^2=25, so we get the points (0, 5) & (0, -5)
if y=0, the equation gets transformed into x^2= 25, so we get the points (5,0) & (-5, 0)

It is not finished yet, you have to ask yourself if there are other possible combinations of x & y as integers mantaining the value of the equation. Sure there are 9 + 16 = 25, same as 16+ 9 =25.

so x can be 3 or -3. Y can be 4 or -4.
simingly x can be -4 or 4 and y can be -3 or 3.
That is how the other point (3, 4), (-3, 4), (3, -4), (-3, -4), (4,3), (4, -3), (-4, 3), (-4, -3).

so at the end there are 12 points with integer coordinates.
riba made
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by fskilnik@GMATH » Fri Nov 12, 2010 11:57 am
vaibhavjha wrote:If the radius of a circle that centers at origin is 5. How many points on the circle have integer coordinates?

a.4
b.8
c.12
d.16
e.20
It seems to me that the "issue" here is to EASILY guarantee that we are not missing any (x,y) point with integer coordinates such that x^2 + y^2 = 25, right?

To be able to do that, please note that every (x,y) in the condition above is such that x^2 and y^2 are PERFECT SQUARES, therefore we are able to find the answer EASILY AND SAFELY through that:

> If x^2 = 5^2 = 25 and y^2 = 25-25 = 0,
that means x = 5 or -5, y = 0, so we have two points: (5,0) and (-5,0).

> If x^2 = 4^2 = 16 and y^2 = 25 - 16 = 9,
that means x = 4 or -4, y = 3 or -3, so we have four points: (4,3), (-4,3), (4, -3) and (-4, -3).

> If x^2 = 3^2 = 9 and y^2 = 25 - 9 = 16,
that means four additional points (who are they?)

> If x^2 = 2^2 = 4 and y^2 = 25 - 4 = 21
we have no points, because 21 is not a perfect square.

> If x^2 = 1^2 = 1 and y^2 = 25 - 1 = 24
we have no points, because 24 is not a perfect square.

> If x^2 = 0^2 = 0 and y^2 = 25 - 0 = 25
we have two additional points (who are they?)

There we are, 2* (4+2) = 12 points, with a bullet-proof trivial method to guarantee no points are missing and no points were counted more than once.

I guess it cannot be easier (and mathematically perfectly justified), can it?

Regards,
Fabio.
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by Target2009 » Fri Nov 12, 2010 8:10 pm
IMO - 12 ( C)
1. 4 point on XY axis.
2.
X^2 + Y^2 = 5^2
(X,Y) = 3,2 or 2,3
in all 4 quadrant do 8 points for (3,2) (2,3) combination

Total = 4 + 8 = 12 points
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by thebigkats » Sun Dec 19, 2010 11:20 am
Hi:
I took the rookie path:
Draw a circle in your min with (0,0) at center. Now draw a line from any x,y to center and you see that the this line is Radius. So as other suggested it is easy to come to conclusion that x^2+y^2=R^2=25

So then I went from one point on x axis to another quickly to see that:

1. x=-5 --> y=0
2. x=-4 --> y=3/-3
3. x=-3 --> y=4/-4
4. x=-2--> y=sqrt(21) --> non integer
5. x=-1 --> y =sqrt(-24) --> non integer
6. x=0 --> y=5/-5
7-11. x=1..5 -->same as x=-1..-5 --> will resolve for x=3 (y=4/-4), 4 (y=3/-3),5 (y=0)

So total = 5*2 + 2 = 12
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by anshumishra » Sun Dec 19, 2010 12:36 pm
x^2+y^2 = 25

What are the possible integral solutions for x ( and also for y) :

([0,+-5],[+-3,+-4])

In terms of x-y co-ordinates :

(0,5),(5,0),(0,-5),(-5,0),(3,4),(4,3),(3,-4),(-4,3),(-3,4),(4,-3),(-3,-4),(-4,-3)

Hence total number of points = 12.
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by Zerks87 » Thu Dec 30, 2010 11:41 am
Rahul@gurome wrote:Let any point on the circle be denoted by (x,y).
Since centre of the circle is origin (0, 0) and radius is 5, we get that x^2 + y^2 = 5^2 = 25.
So the possible values of (x, y) are (3, 4), (3, -4), (-3, 4), (-3, -4), (4, 3), (4, -3), (-4, 3), (-4, -3), (5, 0), (-5, 0), (0, 5), (0, -5) where x and y can be integers only.
There are in total 12 points.

The correct answer is c.
I was wondering why with this question do we square both the x and y coordinate. I understand the fact that all 5,0 combinations are points on the circle and understand that if 5 squared = 25 then all combinations where the sum of the two squared coordinated equals 25 would also be on the circle. I am just not clear as to why you square the x,y coordinates in the first place.
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by ragz » Thu Dec 30, 2010 1:08 pm
Zerks87 wrote:
Rahul@gurome wrote:Let any point on the circle be denoted by (x,y).
Since centre of the circle is origin (0, 0) and radius is 5, we get that x^2 + y^2 = 5^2 = 25.
So the possible values of (x, y) are (3, 4), (3, -4), (-3, 4), (-3, -4), (4, 3), (4, -3), (-4, 3), (-4, -3), (5, 0), (-5, 0), (0, 5), (0, -5) where x and y can be integers only.
There are in total 12 points.

The correct answer is c.
I was wondering why with this question do we square both the x and y coordinate. I understand the fact that all 5,0 combinations are points on the circle and understand that if 5 squared = 25 then all combinations where the sum of the two squared coordinated equals 25 would also be on the circle. I am just not clear as to why you square the x,y coordinates in the first place.
The reason we are squaring is because if there is such a point with integer coordinates on the circle, supposedly (x,y), they should be at a distance of the radius (5) from the origin. And, since the formula for distance between two points is:

(x1 - x2)^2 + (y1 - y2)^2 = (distance)^2

==> (x - 0)^2 + (y - 0)^2 = (5)^2
==> x^2 + y^2 = 25.

I hope this helps understand why this equation will give you the answer.

--
Ravi.
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by mayank82 » Mon Sep 26, 2011 7:15 am
This might help to newbies like me
Equation of a Circle
(x-a)^2 + (y-a)^2 = r^2
where a and b are the coordinates of the origin and r is the radius.
since center of the circle is the origin that means x^2 + y^2 = 5^2

question requires us to find the integer cordinates on the circumfrence of the circle
apply pyth. theorum
x^2 + y^2 = 5^2
this euqation is satisfied by (3,4) (4,3) (0,5) (5,0). Now (3,4) and (4,3) repeats themselves in each coordinate and there are a total of 4 quads so 4*2 =8
(0,5) will have a -ve x equivalent (0,5) and (0,-5) =2
(5,0) ------------do------------- (5,0) and (-5,0) =2
total =12
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by immaculatesahai » Fri Dec 16, 2011 12:22 am
C wins.

Just remember that apart from (0,5),(0,-5),(5,0) and (-5,0) you will also see variants of 3 and 4, because 3,4,5 form the pythagoran triplet.

Just plot the points on a figure and it is easy to figure out.
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by ankush123251 » Fri Dec 16, 2011 2:01 am
Equation of circle with centre at origin is :
x^2 + y^2 = r^2= 5^2 = 25...

Dont care about anything just find the number of points on circle that satisfy above equation in first quadrant.
As you will see,(5,0),(3,4),(4,3) are the coordinates.
Simply multiply by 4 as symmetry will be same for rest of quadrants.
So answer is 3*4 =12.

PS:@Mayank82 the formula is correct but i believe a and b are the coordinates of the center of the circle and not the origin.
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