Find how many points on a circle have an integer distance from other points

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SUMMARY

This discussion focuses on determining how many points on a circle have an integer distance from a specific point, (-4, 5). The distance formula is derived as $$d=\sqrt{12x-16y+53}$$ after substituting the circle's equation. Participants suggest changing coordinates to simplify calculations and utilizing symmetry to limit the number of points that need to be checked. The consensus is that the answer must be an even number, and only two points on the circle need to be evaluated for integer distances.

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  • Understanding of the distance formula in a Cartesian coordinate system
  • Knowledge of circle equations and the concept of radius
  • Familiarity with coordinate transformations, specifically shifting the origin
  • Basic understanding of symmetry in geometric problems
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songoku
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Homework Statement
Please see below
Relevant Equations
##(x-a)^2+(y-b)^2=r^2##

Distance between 2 points
1652872278696.png


Distance between point (-4, 5) and point on circle:
$$d=\sqrt{(x+4)^2+(y-5)^2}$$
$$=\sqrt{x^2+8x+16+y^2-10y+25}$$

Then substitute ##y^2## from equation of circle:
$$d=\sqrt{x^2+8x+16-x^2+4x-6y+12-10y+25}$$
$$=\sqrt{12x-16y+53}$$

After this, I need to try the points one by one to check whether the distance is an integer and also whether the point is located on the circle? I am pretty sure there should be a more sane method to do this

Thanks
 
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I would start by changing coordinates to put the origin at circle center. Then there are only a few right triangles with integer sides. Actually I'll bet you can show it must be an even number,which is all you need for this.
More later if necessary .
 
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Thank you very much hutchphd
 
(Inspired from @hutchphd ’s reply):

Can you workout the center of the circle? We would get it as (2,-3). And radius?

Take any point on the circle, draw a triangle joining the center, the point (-4,5) and the arbitrary point on the circle (x,y). We know the distance between the point on circle and the center is ##r## (please workout it out) and the distance between the center and (-4,5) is 10 units. With resepct to the angle made at the center in this triangle, can you write out the cosine rule?
 
Hall said:
With resepct to the angle made at the center in this triangle, can you write out the cosine rule
Or just "complete the square" to find the original radius.
I seem to be having trouble getting any of the prescribed answers although my symmetry argument seems correct. But I have been wrong before!
 
First, a change of coordinates to ##\bar x = x - 2, \ \bar y = y + 3## is a good idea.

Then use the symmetry to move the point ##P## to somewhere simpler, keeping the same relationship with the circle.

Then count.

However, as the circle itself has an integer radius ...
 
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It does? Jeez I must have screwed up. I wish I could do arithmetic.
 
Last edited:
songoku said:
1652872278696-png.png


After this, I need to try the points one by one to check whether the distance is an integer and also whether the point is located on the circle? I am pretty sure there should be a more sane method to do this

Thanks
No, you only need to check the distance for at most 2 points on the circle. Those being the point nearest to ##(-4,~5)## and the one farthest from ##(-4,~5)##.

How many integers fall between those two extremes?
 
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SammyS said:
No, you only need to check the distance for at most 2 points on the circle. Those being the point nearest to (−4, 5) and the one farthest from (rs by symmetry−4, 5).
Except I believe the points (almost) always come in equidistant pairs...
 
  • #10
hutchphd said:
Except I believe the points (almost) always come in equidistant pairs...
I don't claim that's the entire solution.

Let's leave something for song-o to figure out.

(Added later, just so you don't think I'm time traveling)

Yes, I also get one of the listed answers.
 
Last edited:
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  • #11
Did you get one of the multiple listed answers (we can do that much)?
 
  • #12
hutchphd said:
Did you get one of the multiple listed answers (we can do that much)?
Yes.
 
  • #13
I will look forward to the complete answer at the appropriate time ...!

EDIT DUH!: When you complete the square the quadratic terms are always positive. I wish I could do arithmetic. Thanks.
 
  • #14
I got the answer using @hutchphd hint in post#2 but the method I use is not as efficient as @SammyS in post#8

But sorry I can't continue the hint given by @PeroK in post#6
 
  • #15
songoku said:
But sorry I can't continue the hint given by @PeroK in post#6
By symmetry, the answer must be an even number.
 
  • #16
PeroK said:
By symmetry, the answer must be an even number.
If all the options were even numbers, how to move point P to obtain the answer?

Thanks
 
  • #17
songoku said:
If all the options were even numbers, how to move point P to obtain the answer?
Rotate the system to put ##P## on the ##x##-axis.
 
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  • #18
Thank you very much Hall, hutchphd, SammyS, PeroK
 
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  • #19
To calculate the answer you need the radius of the circle and the distance between P and the center of the circle. After that the coordinates are irrelevant.
 
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  • #20
Since the problem appears solved, here are some possibly interesting visualizations of the problem.
(Try the "interesting views" at the bottom.)

https://www.desmos.com/calculator/cf30ycwkta

There's a hint of a related physics problem: interference from two coherent point sources.
 
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