A circle has center (7,5) and radius 5, if (x,y) is a point on the circle, how many such points on the circle are possible? (x and y both are integers)
Please explain this question in details.
GRE Forum Archive
6 postsPage 1 of 1
- md.sharifulislam01915
- Students
- Posts: 7
- Joined: Sun Dec 08, 2013 3:38 pm
- tommywallach
- Manhattan Prep Staff
- Posts: 1917
- Joined: Thu Mar 31, 2011 11:18 am
Re: Circle
Hey MD,
The answer will be 12. Where did you find this question? It seems a little hard for the GRE.
Basically, coordinate graph questions are always about either line equations or right triangles. This question is pretending it's about circles, but it isn't really. Think of it this way.
4 of the points will simply be the points you get when you go up 5, left 5, right 5, or down 5 from the center point.
The other 8 are a little more complex. Remember that the distance between any two points in a coordinate graph can always be expressed as a right triangle (try drawing a line between any two points). Then you can build a right triangle with the difference between the X values and the Y values of the two points as the length of the two perpendicular sides.
Because this circle has a radius of 5, it will create triangles of 3-4-5 and 4-3-5 (in one case, the side that is parallel to the x-axis is 3 and the side that is parallel to the y-axis is 4; in the other case, the side that is parallel to the x-axis is 4 and the side that is parallel to the y-axis is 3). It will do this in each of the four "quadrants" of the circle, making for 8 more right triangles with integer side values.
Hope that helps!
-t
The answer will be 12. Where did you find this question? It seems a little hard for the GRE.
Basically, coordinate graph questions are always about either line equations or right triangles. This question is pretending it's about circles, but it isn't really. Think of it this way.
4 of the points will simply be the points you get when you go up 5, left 5, right 5, or down 5 from the center point.
The other 8 are a little more complex. Remember that the distance between any two points in a coordinate graph can always be expressed as a right triangle (try drawing a line between any two points). Then you can build a right triangle with the difference between the X values and the Y values of the two points as the length of the two perpendicular sides.
Because this circle has a radius of 5, it will create triangles of 3-4-5 and 4-3-5 (in one case, the side that is parallel to the x-axis is 3 and the side that is parallel to the y-axis is 4; in the other case, the side that is parallel to the x-axis is 4 and the side that is parallel to the y-axis is 3). It will do this in each of the four "quadrants" of the circle, making for 8 more right triangles with integer side values.
Hope that helps!
-t
- md.sharifulislam01915
- Students
- Posts: 7
- Joined: Sun Dec 08, 2013 3:38 pm
Re: Circle
Hey tommywallach,
I got this question from one of my friend who recently participated in the gre exam and got this similar type of question. I am not understanding it clearly. It is helpful for me if you draw a figure with explanation.
I got this question from one of my friend who recently participated in the gre exam and got this similar type of question. I am not understanding it clearly. It is helpful for me if you draw a figure with explanation.
- tommywallach
- Manhattan Prep Staff
- Posts: 1917
- Joined: Thu Mar 31, 2011 11:18 am
Re: Circle
I've never seen a question this difficult on the GRE, so I don't believe it (I know this as a GMAT question). However, I'll try to explain it without pictures (I don't have a paint program!).
Do this:
1. Draw a circle (no coordinate graph necessary).
2. Separate the circle into 4 equal pieces (by drawing a horizontal and a vertical diameter).
3. Draw a line from the center of the circle to a location on the circumference about 1/3 of the way between the topmost point of the circle and the rightmost point of the circle.
4. That length is 5 (obviously, it's a radius).
5. Draw a line down from that point to the horizontal diameter of the circle that you've drawn. You now have a right triangle. Imagine that the sides of this triangle are 3 and 4, creating a 3-4-5 right triangle. This triangle exists. Obviously, you didn't actually draw it (because you just guessed at that top point), but it EXISTS. All three points of this triangle have integer values.
6. Now, draw a line from the center of the circle to a point on the circumference about 2/3 of the way between the topmost point of the circle and the rightmost point of the circle. Imagine that this is another 3-4-5 triangle, only the height of this triangle is 3 (whereas the height of the other triangle was 4). This triangle also exists.
7. These two 3-4-5 triangles exist in all 4 quadrants.
Make sense?
-t
Do this:
1. Draw a circle (no coordinate graph necessary).
2. Separate the circle into 4 equal pieces (by drawing a horizontal and a vertical diameter).
3. Draw a line from the center of the circle to a location on the circumference about 1/3 of the way between the topmost point of the circle and the rightmost point of the circle.
4. That length is 5 (obviously, it's a radius).
5. Draw a line down from that point to the horizontal diameter of the circle that you've drawn. You now have a right triangle. Imagine that the sides of this triangle are 3 and 4, creating a 3-4-5 right triangle. This triangle exists. Obviously, you didn't actually draw it (because you just guessed at that top point), but it EXISTS. All three points of this triangle have integer values.
6. Now, draw a line from the center of the circle to a point on the circumference about 2/3 of the way between the topmost point of the circle and the rightmost point of the circle. Imagine that this is another 3-4-5 triangle, only the height of this triangle is 3 (whereas the height of the other triangle was 4). This triangle also exists.
7. These two 3-4-5 triangles exist in all 4 quadrants.
Make sense?
-t
- md.sharifulislam01915
- Students
- Posts: 7
- Joined: Sun Dec 08, 2013 3:38 pm
Re: Circle
Hey tommywallach,
Thanks for your help.
Thanks for your help.
- tommywallach
- Manhattan Prep Staff
- Posts: 1917
- Joined: Thu Mar 31, 2011 11:18 am
Re: Circle
Glad to help! Hope it made sense without a picture! : )
-t
-t
6 posts Page 1 of 1