The set P contains the points (x, y) on the coordinate plane that are

Question

The set P contains the points (x, y) on the coordinate plane that are in or on circle O. The values of x and y are integers. Circle O is centered at the origin and has a radius of 3. If a point from set P is randomly selected, what is the probability that the point is located on the circumference of circle O?

A. 4/29
B. 4/28
C. 4/27
D. 4/19
E. 2/9

Kudos for a correct solution.

A. 4/29
B. 4/28
C. 4/27
D. 4/19
E. 2/9

chetan2u · Accepted Answer

Hi,
the simplest way to solve the Q here is as follows...
1)we know there are four sets of integers on the circumference... so the numerator is 4..
now we should have the denominator in the form 4X +1... only 29 fits in so ans 4/29..
although 9 is there but it will become 18 when num is 4..
 WHY 4X+1? we know there are four quadrants, and there will be total 4x sets where x is total numbers in one Quad  AND ADD TO IT THE SINGLE ORIGIN (0,0) 
ans A..

2)if the choices are not so obvious ..
0,0 -1
0,1-2
0,2-4
0,3-4
1,0-2
1,1-4
1,2-4
2,1-4
2,2-4
total 4*7+1=29
ans A

balamoon · Answer

Solution -

Circle O with center at the origin and radius 3. Since x and y are integers, the circle touches the x and y axes at four points (3, 0), (0, 3), (-3, 0) and (0, -3).

Number of points inside circle O in the first quadrant are (1, 1), (1, 2), (2, 1) and (2, 2). Using symmetry, each of the 4 quadrants has 4 points. So there are 4 × 4 = 16 points inside the quadrants.

Now look at the axes. The x-axis has the 7 points (-3, 0), (-2, 0), (-1, 0), (0, 0), (1, 0), (2, 0), (3, 0). The y-axis also has 7 points. But both axes have counted the origin, so the sum is one less. So there are 7 + 7 - 1 = 13 points on the axes.

The total number of points inside and on the circle is 16 + 13 = 29.

From the graph attached, there are 4 points on the circumference of the circle. So the probability is 4/29.

ENGRTOMBA2018 · Answer

The standard equation of circle with center (a,b) and radius = r is  (x-a)^2+(y-b)^2=r^2

For a circle at the center, (a,b) = (0,0), radius = 3

Thus the equation becomes ,  x^2+y^2 = 9

Now all the possible points (all integers only) inside and on the circumference (in red) are  x^2+y^2<9 , and for points on circumference (x,y) will satisfy  x^2+y^2=9 ] :

(0,0)
(1,0)
(2,0)
 (3,0) 
(-1,0)
(-2,0)
 (-3,0) 
(0,1)
(0,2)
(0,-1)
(0,-2)
 (0,3)
(0,-3) 
(-2,-1)
(-2,-2)
(-2,1)
(-2,2)
(-1,-1)
(-1,-2)
(-1,1)
(-1,2)
(2,-1)
(2,-2)
(2,1)
(2,2)
(1,-1)
(1,-2)
(1,1)
(1,2)

Thus we see , total points = 29, on circumference = 4

Thus the probability = 4/29. A is the correct answer.

GMATinsight · Answer

1) We don't necessarily have to find out all points in all Quadrants. No. of Points in Every Quadrant will be same so we need to calculate only the No. of points in one quadrant and multiply it by 4

2) There are 4 Axes( + X and  + Y and we must take one axes with one quadrant

3) The point at the origin must be considered separately at the end as Origin involves all quadrants

Answer Option A

Bunuel · Answer

800score Official Solution:

Graph circle O with center at the origin and radius 3. The circle touches the x. and y axes at the four points (3, 0), (0, 3), (-3, 0) and (0, -3).

Since x and y are integers, it is easy to see and count the number of points inside circle O. In the first quadrant, the points are (1, 1), (1, 2), (2, 1) and (2, 2). Using symmetry, each of the 4 quadrants has 4 points. So there are 4 × 4 = 16 points inside the quadrants.

Now look at the axes. The x-axis has the 7 points (-3, 0), (-2, 0), (-1, 0), (0, 0), (1, 0), (2, 0), (3, 0). The y-axis also has 7 points. But both axes have counted the origin, so the sum is one less. So there are 7 + 7 - 1 = 13 points on the axes.
The total number of points inside and on the circle is 16 + 13 = 29.

From the graph, there are 4 points on the circumference of the circle. So the probability is 4/29.

shashankism · Answer

At the first instance it looks a very complex or calculation intensive problem. But if we solve some of the similar problems we will easily understand how to tackle such problems easily.

Well lets start with the coordinates which lies on the circle and we know that these points are (0,3), (3,0),(0,-3),(-3,0) (Just check points anticlockwise though its not mandatory.. )

Now lets start finding all the integer coordinates either inside the circle or on the circle.
We will first check points in the first quadrant which doesn't lie on the x axis or y -axis as the points on a&y axes are common to all quadrants.
So, in the first quadrant points are (1,1)(1,2)(2,1)(2,2) ... Total 4

So, No. of points which has integral coordinates and lies within/on the circle and does not lie on x or y axis = 4*4 = 16

Now lets start finding all the integer coordinates on the positive x- axis excluding origin 
Such point are (1,0)(2,0), (3,0) .. Total 3 
So, No. of points which has integral coordinates and lies within/on the circle and lie on x or y axis (excluding origin)= 3*4 = 12

Origin ..... Total 1

So, Total no. of points which has integral coordinates and lies within/on the circle = 16+12+1 = 29

probability = 4/29

Answer A..

This may be represented by a diagram also as provided by many users above.. You can refer the same... Thanks

Adarsh_24 · Answer

I got this initially wrong.
x^2+y^2>=9
Possible integers 0,1,2,3,-1,-2,-3.
Assume x=0 then y can be 0,1,2,3,-1,-2,-3 => 7 combos
x=1 then y can be 0,1,2,-1,-2 => 5 combos
x=2 then y can be 0,1,2,-1,2 => 5 
X=3 then y can be 0 only => 1
So total = 7+5*2+5*2+1*2=29
Multiplied by 2 to account for x=-1,-2,-3
On the circle x^2+y^2=9 exact.
There are four (3,0), (-3,0), (0,-3), (0,3).
Ans 4/29

But, chetan2u method is the best.

The set P contains the points (x, y) on the coordinate plane that are

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