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If the radius of a circle that centers at the origin is 5, how many po
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21 Jan 2015, 02:39
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If the radius of a circle that centers at the origin is 5, how many points on the circle have integer coordinates? (A) 4 (B) 8 (C) 12 (D) 15 (E) 20 Kudos for a correct solution.
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Re: If the radius of a circle that centers at the origin is 5, how many po
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02 Feb 2015, 05:16
Bunuel wrote: If the radius of a circle that centers at the origin is 5, how many points on the circle have integer coordinates?
(A) 4 (B) 8 (C) 12 (D) 15 (E) 20
Kudos for a correct solution. There is a necessity to understand some of the verbiage in this question in order to be able to answer it properly. Firstly, a circle that is centered at the origin is centered at point {0,0}. The radius is 5, which means we know the diameter (2*r), the circumference (2*π*r) and the area (π * r^2). However, none of that information really helps us to answer this question. We are interested in how many points on the circle have integer coordinates. Quite simply, a circle has an infinite number of discrete points, so it’s easier to answer this question in the reverse: For each integer coordinate, is that point on the circle? Let’s start with the obvious ones. The point {5,0} has to necessarily be on the circle. If the origin is {0,0} and the radius is 5, then not only must point {5,0} be on the circle, but so must point {5,0}. The circle extends in all four directions, so we cannot forget the negative values. Similarly, the points {0,5} and {0,5} will also be on the points, effectively covering the four cardinal points from the original circle. The circle could look something like this: Attachment:
Geometry_Figure1.jpg [ 8 KiB  Viewed 2904 times ]
After solving for these four points, we must evaluate whether other integer coordinates could be on the circle. One thing should be clear: if the radius is 5, then any integer point above 5 will necessarily not be on the circle, as it is beyond the reach of our radius. We’ve already covered zero, so the only options we have left are one, two, three and four. Of course all of these numbers have negatives and can be considered on either the x or y axis, but still we have a finite number of possibilities to consider. Another important thing that might not be as obvious is that the answer to this question will necessarily be a multiple of four. Given that a circle extends in all directions by the same distance, it is impossible for point {x, y} to be on the circle and for points {x,y}, {x,y} and {x,y} to not also be on the circle. This is an important property of all circles and one of the reasons they’re so common in everything from architecture to cooking (and to alien crop circles, if you believe in that). This rule also guarantees that any answer choice that’s not a multiple of four can be eliminated. We can thus eliminate answer choice D (15). How do we go about finding other points on the circle? (Why am I asking rhetorical questions?) By using the Pythagorean Theorem, of course! Any point on the circle naturally forms a right angle triangle with the radius as the hypotenuse, and the radius is always five. Therefore, if the two other sides can be formed out of integers, we have a point on the circle with integer coordinates. The graph below will highlight this principle: Attachment:
Geometry_Figure2.jpg [ 7.4 KiB  Viewed 2905 times ]
Since the Pythagorean Theorem states that the squares of the sides will be equal to the square of the hypotenuse, we only need to look for numbers that satisfy the equation a^2 +b^2 = r^2. And given that r is 5, r^2 must always be 25. So if we plug in a as one, we find that 1 + b^2 = 25. This gives us b^2 = 24, or b = √24, which is not an integer. We only have to plug this in three more times, so there’s no reason not to try all the possibilities. If a = 2, then we get 4 + b^2 = 25. The value of b would be √21, which again is not an integer value. If a = 3, however, we quickly recognize the vaunted 345 triangle, as 9 + b^2 = 25, meaning b^2 is 16 and therefore b is 4. This means that the points {3,4}, {3,4}, {3,4} and {3,4} are all on the circle. We’ve brought the total up to 8, but we’re not done. The final value is when a equals four, which will again work and bring in the converse of the last iteration: {4,3}, {4,3}, {4,3} and {4,3}. These values are distinct from the previous ones, so we now have a total of 12 points. We’ve already checked five, so we can stop here. The answer to this question is answer choice C. There will be 12 distinct values with integer coordinates, as crudely demonstrated below (or on any analog clock). Attachment:
Geometry_Figure3.jpg [ 14.45 KiB  Viewed 2904 times ]
In geometry, even if it feels like you have to constantly commit more rules to memory, remember that the rules are not nearly as important as knowing how to apply them. This problem can be solved with just the Pythagorean Theorem and a little elbow grease (or a graphing calculator). The GMAT is very much a test of how you think, not of what you know. If you think about geometry problems as cases that must be solved, or obstacles to be overcome, you’ll be in good shape to solve them.
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Re: If the radius of a circle that centers at the origin is 5, how many po
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21 Jan 2015, 06:08
My thoughts: if you have a radius of 5 you automatically know that there will be four points 5,0 , 0,5 , 5,0, 0,5.
Now you know the distance to any point on the circle is five. Let's use the distance formula. You know your x coordinates can only be 14.
D^2 = 25 so now square all your x coordinates and see which values when subtracted from 25 will give you a perfect. The x values that work are 3, 4, 3, 4. So you'll have another 4 points.



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Re: If the radius of a circle that centers at the origin is 5, how many po
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Updated on: 23 Jan 2015, 21:00
12.... the radius is 5 so it becomes the hypotenuse of right angle triangle with X & Y as other two sides. only sides of (3,4)&(4,3) and (0,5)(a straight line in these cases) including the ve values so 2 value in each quadrant and 4 on axis
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Originally posted by chetan2u on 21 Jan 2015, 07:01.
Last edited by chetan2u on 23 Jan 2015, 21:00, edited 1 time in total.



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Re: If the radius of a circle that centers at the origin is 5, how many po
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23 Jan 2015, 14:33
Hi All, It looks like everyone has properly recognized the axis points (5,0), (0,5), (5,0) and (0,5) and noticed that the radius (of length 5) creates the hypotenuse of a right triangle when placed as a diagonal into the graph. Since we're restricted to INTEGER coordinates, we're looking for 3s and 4s (to complete the 3/4/5 right triangles that can be formed). However, you have to be thorough and "see" ALL of the possibilities. There are more than you realize.... (3,4) and (4,3) (3,4) and (4,3) (3,4) and (4,3) (3,4) and (4,3) ....and the 4 axis points. This gives us a total of 12 possibilities. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If the radius of a circle that centers at the origin is 5, how many po
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03 Mar 2016, 21:25
Bunuel wrote: If the radius of a circle that centers at the origin is 5, how many points on the circle have integer coordinates?
(A) 4 (B) 8 (C) 12 (D) 15 (E) 20
Kudos for a correct solution. I understand this might not be required but I used the equation of a circle. Since the origin is at 0, x^2+y^2=5^2. X,Y could be +/ (0,5 or 5,0)  4 possibilities. X,Y could be +/ (3,4 or 4,3)  8 possibilities. Ans: C



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Re: If the radius of a circle that centers at the origin is 5, how many po
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11 Jun 2017, 10:35
Bunuel wrote: Bunuel wrote: If the radius of a circle that centers at the origin is 5, how many points on the circle have integer coordinates?
(A) 4 (B) 8 (C) 12 (D) 15 (E) 20
Kudos for a correct solution. There is a necessity to understand some of the verbiage in this question in order to be able to answer it properly. Firstly, a circle that is centered at the origin is centered at point {0,0}. The radius is 5, which means we know the diameter (2*r), the circumference (2*π*r) and the area (π * r^2). However, none of that information really helps us to answer this question. We are interested in how many points on the circle have integer coordinates. Quite simply, a circle has an infinite number of discrete points, so it’s easier to answer this question in the reverse: For each integer coordinate, is that point on the circle? Let’s start with the obvious ones. The point {5,0} has to necessarily be on the circle. If the origin is {0,0} and the radius is 5, then not only must point {5,0} be on the circle, but so must point {5,0}. The circle extends in all four directions, so we cannot forget the negative values. Similarly, the points {0,5} and {0,5} will also be on the points, effectively covering the four cardinal points from the original circle. The circle could look something like this: Attachment: Geometry_Figure1.jpg After solving for these four points, we must evaluate whether other integer coordinates could be on the circle. One thing should be clear: if the radius is 5, then any integer point above 5 will necessarily not be on the circle, as it is beyond the reach of our radius. We’ve already covered zero, so the only options we have left are one, two, three and four. Of course all of these numbers have negatives and can be considered on either the x or y axis, but still we have a finite number of possibilities to consider. Another important thing that might not be as obvious is that the answer to this question will necessarily be a multiple of four. Given that a circle extends in all directions by the same distance, it is impossible for point {x, y} to be on the circle and for points {x,y}, {x,y} and {x,y} to not also be on the circle. This is an important property of all circles and one of the reasons they’re so common in everything from architecture to cooking (and to alien crop circles, if you believe in that). This rule also guarantees that any answer choice that’s not a multiple of four can be eliminated. We can thus eliminate answer choice D (15). How do we go about finding other points on the circle? (Why am I asking rhetorical questions?) By using the Pythagorean Theorem, of course! Any point on the circle naturally forms a right angle triangle with the radius as the hypotenuse, and the radius is always five. Therefore, if the two other sides can be formed out of integers, we have a point on the circle with integer coordinates. The graph below will highlight this principle: Attachment: Geometry_Figure2.jpg Since the Pythagorean Theorem states that the squares of the sides will be equal to the square of the hypotenuse, we only need to look for numbers that satisfy the equation a^2 +b^2 = r^2. And given that r is 5, r^2 must always be 25. So if we plug in a as one, we find that 1 + b^2 = 25. This gives us b^2 = 24, or b = √24, which is not an integer. We only have to plug this in three more times, so there’s no reason not to try all the possibilities. If a = 2, then we get 4 + b^2 = 25. The value of b would be √21, which again is not an integer value. If a = 3, however, we quickly recognize the vaunted 345 triangle, as 9 + b^2 = 25, meaning b^2 is 16 and therefore b is 4. This means that the points {3,4}, {3,4}, {3,4} and {3,4} are all on the circle. We’ve brought the total up to 8, but we’re not done. The final value is when a equals four, which will again work and bring in the converse of the last iteration: {4,3}, {4,3}, {4,3} and {4,3}. These values are distinct from the previous ones, so we now have a total of 12 points. We’ve already checked five, so we can stop here. The answer to this question is answer choice C. There will be 12 distinct values with integer coordinates, as crudely demonstrated below (or on any analog clock). Attachment: Geometry_Figure3.jpg In geometry, even if it feels like you have to constantly commit more rules to memory, remember that the rules are not nearly as important as knowing how to apply them. This problem can be solved with just the Pythagorean Theorem and a little elbow grease (or a graphing calculator). The GMAT is very much a test of how you think, not of what you know. If you think about geometry problems as cases that must be solved, or obstacles to be overcome, you’ll be in good shape to solve them. Thanks for the detailed solution on this. This was very helpful. I have a question here. Lets say if there is a twist in the question and question is about inside the circle rather than on the circle. Is there a easy way to solve this? Or it is OOS for GMAT



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Re: If the radius of a circle that centers at the origin is 5, how many po
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11 Jun 2017, 11:48
vnitnagpur wrote: Bunuel wrote: Bunuel wrote: If the radius of a circle that centers at the origin is 5, how many points on the circle have integer coordinates?
(A) 4 (B) 8 (C) 12 (D) 15 (E) 20
Kudos for a correct solution. There is a necessity to understand some of the verbiage in this question in order to be able to answer it properly. Firstly, a circle that is centered at the origin is centered at point {0,0}. The radius is 5, which means we know the diameter (2*r), the circumference (2*π*r) and the area (π * r^2). However, none of that information really helps us to answer this question. We are interested in how many points on the circle have integer coordinates. Quite simply, a circle has an infinite number of discrete points, so it’s easier to answer this question in the reverse: For each integer coordinate, is that point on the circle? Let’s start with the obvious ones. The point {5,0} has to necessarily be on the circle. If the origin is {0,0} and the radius is 5, then not only must point {5,0} be on the circle, but so must point {5,0}. The circle extends in all four directions, so we cannot forget the negative values. Similarly, the points {0,5} and {0,5} will also be on the points, effectively covering the four cardinal points from the original circle. The circle could look something like this: Attachment: Geometry_Figure1.jpg After solving for these four points, we must evaluate whether other integer coordinates could be on the circle. One thing should be clear: if the radius is 5, then any integer point above 5 will necessarily not be on the circle, as it is beyond the reach of our radius. We’ve already covered zero, so the only options we have left are one, two, three and four. Of course all of these numbers have negatives and can be considered on either the x or y axis, but still we have a finite number of possibilities to consider. Another important thing that might not be as obvious is that the answer to this question will necessarily be a multiple of four. Given that a circle extends in all directions by the same distance, it is impossible for point {x, y} to be on the circle and for points {x,y}, {x,y} and {x,y} to not also be on the circle. This is an important property of all circles and one of the reasons they’re so common in everything from architecture to cooking (and to alien crop circles, if you believe in that). This rule also guarantees that any answer choice that’s not a multiple of four can be eliminated. We can thus eliminate answer choice D (15). How do we go about finding other points on the circle? (Why am I asking rhetorical questions?) By using the Pythagorean Theorem, of course! Any point on the circle naturally forms a right angle triangle with the radius as the hypotenuse, and the radius is always five. Therefore, if the two other sides can be formed out of integers, we have a point on the circle with integer coordinates. The graph below will highlight this principle: Attachment: Geometry_Figure2.jpg Since the Pythagorean Theorem states that the squares of the sides will be equal to the square of the hypotenuse, we only need to look for numbers that satisfy the equation a^2 +b^2 = r^2. And given that r is 5, r^2 must always be 25. So if we plug in a as one, we find that 1 + b^2 = 25. This gives us b^2 = 24, or b = √24, which is not an integer. We only have to plug this in three more times, so there’s no reason not to try all the possibilities. If a = 2, then we get 4 + b^2 = 25. The value of b would be √21, which again is not an integer value. If a = 3, however, we quickly recognize the vaunted 345 triangle, as 9 + b^2 = 25, meaning b^2 is 16 and therefore b is 4. This means that the points {3,4}, {3,4}, {3,4} and {3,4} are all on the circle. We’ve brought the total up to 8, but we’re not done. The final value is when a equals four, which will again work and bring in the converse of the last iteration: {4,3}, {4,3}, {4,3} and {4,3}. These values are distinct from the previous ones, so we now have a total of 12 points. We’ve already checked five, so we can stop here. The answer to this question is answer choice C. There will be 12 distinct values with integer coordinates, as crudely demonstrated below (or on any analog clock). Attachment: Geometry_Figure3.jpg In geometry, even if it feels like you have to constantly commit more rules to memory, remember that the rules are not nearly as important as knowing how to apply them. This problem can be solved with just the Pythagorean Theorem and a little elbow grease (or a graphing calculator). The GMAT is very much a test of how you think, not of what you know. If you think about geometry problems as cases that must be solved, or obstacles to be overcome, you’ll be in good shape to solve them. Thanks for the detailed solution on this. This was very helpful. I have a question here. Lets say if there is a twist in the question and question is about inside the circle rather than on the circle. Is there a easy way to solve this? Or it is OOS for GMAT Hi If you take a circle given in the question, x^2 + y^2 = 5^2, the integer points lying on this circle are those that satisfy this equation. However, the points that lie Inside the circle will satisfy the following equation: x^2 + y^2 < 5^2We can rewrite the above equation as: y^2 < 25  x^2Now we will take various integer values of x and check how many integer values of y are possible in each case. We can see that x can only go from 5 to 5 in this case. When x = 5, y^2 < 0, which is not possible. So no integer value here. When x = 5, y^2 is again < 0, which is not possible. So no integer value here also. When x = 4, y^2 < 9, which in turn means 3<y<3. Here y can take 5 integer values (2,1,0,1,2). When x = 4, again y^2 < 9, which in turn means 3<y<3. Here y can again take 5 integer values (2,1,0,1,2). When x = 3, y^2 < 16, which in turn means 4<y<4. Here y can take 7 integer values (3,2,1,0,1,2,3). When x = 3, again y^2 < 16, which in turn means 4<y<4. Here y can again take 7 integer values (3,2,1,0,1,2,3). When x = 2, y^2 < 21. Here y can take 9 integer values (4,3,2,1,0,1,2,3,4). When x = 2, y^2 < 21. Here y can take 9 integer values (4,3,2,1,0,1,2,3,4). When x = 1, y^2 < 24. Here y can take 9 integer values (4,3,2,1,0,1,2,3,4). When x = 1, y^2 < 24. Here y can take 9 integer values (4,3,2,1,0,1,2,3,4). When x = 0, y^2 < 25, which in turn means 5<y<5. Here y can take 9 integer values (4,3,2,1,0,1,2,3,4). Thus total points with integer coordinates lying within the circle are = 5+5+7+7+9+9+9+9+9 = 69. But I think this question is outside the scope of GMAT.



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Re: If the radius of a circle that centers at the origin is 5, how many po
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21 Mar 2019, 15:30
Does this apply to any circle, no matter the origin?



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Re: If the radius of a circle that centers at the origin is 5, how many po
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21 Mar 2019, 22:45
tmunshi wrote: Does this apply to any circle, no matter the origin? The equation of a circle is (x – h)^2 + (y – k)^2 = r^2; where (h,k) is center if Center is Origin then an equation becomes x^2 +y^2=r^2 Hence, it does matter whether a circle is at origin or at any other point.



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Re: If the radius of a circle that centers at the origin is 5, how many po
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22 Mar 2019, 00:34
IMO C
here the radious is 5
so we can have the following points( on circle ) on first quad = 5,0
0,5 4,3 3,4
similarly in second = we have 3
3,4 4,3 5,0
in third = 3 and
3,4 4,3 0,5
the fourth = 2
3,4 4,3
total 12 points
if we apply paythogoras theoram on these points , we get radious 5 award kudos if helpful
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Re: If the radius of a circle that centers at the origin is 5, how many po
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