Re: [Maxima-discuss] trying to solve for intersection of 2 circles

[kostas p. <kos...@ho...>]

From: website.reader via Maxima-discuss<mailto:max...@li...>
Sent: Friday, March 25, 2022 1:37 AM
To: max...@li...<mailto:max...@li...>
Subject: [Maxima-discuss] trying to solve for intersection of 2 circles

I have two circles, first is centered at [1,0] with radius a and the second is at [0,1] with radius b

(1)    (x-1)^2 + y^2 = a^2

(2)    x^2 + (y-1)^2 = b^2

I used some known points, with p = [16/9, 28/15] thus giving a = 91/45 and b = 89/45 for the two circles passing through point p.

When I use maxima to solve for the general case

(%i1) solve([(x-1)^2+y^2-a^2=0,x^2+(y-1)^2-b^2=0],[x,y]);
(%o1)                                 []

Maxima says that the solution is the empty set. However the example showed that we do have a rational solution for known values.

What did I do wrong? If I used the values, Maxima finds 2 intersection points [ -13/15, -7/9] and p = [16/9, 28/15] but it refuses to solve for distances a,b for the general case.

Randall


Hello Randall ,

Since all the circles of radius ‘a’ have their center points in ( 1,0 )
and all the circles of radius ‘b’ in ( 0 , 1 ) , the line Ca Cb  connecting their centers
has always length sqrt(2) . For this case then :

- For every circle of radius ‘a’ so that:  0 < a < sqrt(2)  ,there are infinitely many circles
  of radius ‘b’ so that : b + a < sqrt (2) . Infinite pairs of circles that do not intersect
  and the systems have no solution .
                             a E ( 0 , sqrt (2) ) , b E ( 0 , sqrt ( 2) -a )

-For every circle of radius ‘a’ so that:  0 < a < sqrt(2) , there is always one circle of radius ‘b’
so that : a+b=sqrt(2) .Tangent Circles ,  one common point- the systems have always one
unique solution each time (x,y) .
                               a E (0 ,sqrt (2 ) ) , b = sqrt(2) -a

-For every circle of radius ‘a’ so that:  0 < a < sqrt(2)  ,there are infinitely many circles of radius ‘b’
so that : sqrt(2) -a < b < sqrt(2) + a.   Infinite pairs of circles that always intersect in 2 different
common points ( x,y) . These systems have 2 different solutions ( x, y ) each time .
                              a E ( 0 , sqrt(2) ) , b E ( sqrt(2) -a , sqrt(2)+ a )

-For every circle of radius ‘a’ so that: 0 < a < sqrt(2) there is always one circle of radius ‘b’
so that: b=sqrt(2)+a .Tangent Circles , one common point – Systems have one unique solution
each time ( x, y ) .
                              a E ( 0 , sqrt(2) )  , b = sqrt(2) +a

--For every circle of radius ‘a’ so that:  0 < a < sqrt(2) , there are infinitely many circles of radius ‘b’,
so that : b > sqrt (2) + a . Meaning , infinite pairs of circles that do not intersect ,  and the systems
have no solution .
                               a E ( 0 , sqrt(2) )  , b E ( sqrt(2) +a , +inf )

This is actually elementary theory in geometry , but you can keep that in mind while you formulate
your solution .I do not seem to understand  what you expect as an output from  Maxima or a computer
in general , however I have no doubt that all these could be written in a computer in an elegant way ,
although I cannot help you with that .

You can make also  many other conditions even for this case of Ca ( 1 , 0) and Cb ( 0 ,1 )
but this is generally the case for 2 circles on the 2D plane that have different center points :


  1.  They can either intersect in two common points ,

      2) They can be  tangent to each other internally or externally  sharing one common point ,
or
      3) They do not intersect and do not share common points ,
         -  either sharing some common projection area or not sharing at all- .