Task: Circles of given radius through two points







Python: from collections import namedtuple
from math import sqrt

Pt = namedtuple('Pt', 'x, y')
Circle = Cir = namedtuple('Circle', 'x, y, r')

def circles_from_p1p2r(p1, p2, r):
    'Following explanation at http://mathforum.org/library/drmath/view/53027.html'
    if r == 0.0:
        raise ValueError('radius of zero')
    (x1, y1), (x2, y2) = p1, p2
    if p1 == p2:
        raise ValueError('coincident points gives infinite number of Circles')
    
    dx, dy = x2 - x1, y2 - y1
    
    q = sqrt(dx**2 + dy**2)
    if q > 2.0*r:
        raise ValueError('separation of points > diameter')
    
    x3, y3 = (x1+x2)/2, (y1+y2)/2
    
    d = sqrt(r**2-(q/2)**2)
    
    c1 = Cir(x = x3 - d*dy/q,
             y = y3 + d*dx/q,
             r = abs(r))
    
    c2 = Cir(x = x3 + d*dy/q,
             y = y3 - d*dx/q,
             r = abs(r))
    return c1, c2

if __name__ == '__main__':
    for p1, p2, r in [(Pt(0.1234, 0.9876), Pt(0.8765, 0.2345), 2.0),
                      (Pt(0.0000, 2.0000), Pt(0.0000, 0.0000), 1.0),
                      (Pt(0.1234, 0.9876), Pt(0.1234, 0.9876), 2.0),
                      (Pt(0.1234, 0.9876), Pt(0.8765, 0.2345), 0.5),
                      (Pt(0.1234, 0.9876), Pt(0.1234, 0.9876), 0.0)]:
        print('Through points:\n  %r,\n  %r\n  and radius %f\nYou can construct the following circles:'
              % (p1, p2, r))
        try:
            print('  %r\n  %r\n' % circles_from_p1p2r(p1, p2, r))
        except ValueError as v:
            print('  ERROR: %s\n' % (v.args[0],))




Fortran: 

program circles
implicit none
double precision :: P1(2), P2(2), R

P1 = (/0.1234d0, 0.9876d0/)
P2 = (/0.8765d0,0.2345d0/)
R = 2.0d0
call print_centers(P1,P2,R)

P1 = (/0.0d0, 2.0d0/)
P2 = (/0.0d0,0.0d0/)
R = 1.0d0
call print_centers(P1,P2,R)

P1 = (/0.1234d0, 0.9876d0/)
P2 = (/0.1234d0, 0.9876d0/)
R = 2.0d0
call print_centers(P1,P2,R)

P1 = (/0.1234d0, 0.9876d0/)
P2 = (/0.8765d0, 0.2345d0/)
R = 0.5d0
call print_centers(P1,P2,R)

P1 = (/0.1234d0, 0.9876d0/)
P2 = (/0.1234d0, 0.9876d0/)
R = 0.0d0
call print_centers(P1,P2,R)
end program circles

subroutine print_centers(P1,P2,R)
implicit none
double precision :: P1(2), P2(2), R, Center(2,2)
integer :: Res
call test_inputs(P1,P2,R,Res)
write(*,*)
write(*,'(A10,F7.4,A1,F7.4)') 'Point1  : ', P1(1), ' ', P1(2)
write(*,'(A10,F7.4,A1,F7.4)') 'Point2  : ', P2(1), ' ', P2(2)
write(*,'(A10,F7.4)') 'Radius  : ', R
if(Res.eq.1) then
  write(*,*) 'Same point because P1=P2 and r=0.'
elseif(Res.eq.2) then
  write(*,*) 'No circles can be drawn because r=0.'
elseif(Res.eq.3) then
  write(*,*) 'Infinite circles because P1=P2 for non-zero radius.'
elseif(Res.eq.4) then
  write(*,*) 'No circles with given r can be drawn because points are far apart.'
elseif(Res.eq.0) then
  call find_center(P1,P2,R,Center)
  if(Center(1,1).eq.Center(2,1) .and. Center(1,2).eq.Center(2,2)) then
    write(*,*) 'Points lie on the diameter. A single circle can be drawn.'
    write(*,'(A10,F7.4,A1,F7.4)') 'Center  : ', Center(1,1), ' ', Center(1,2)
  else
    write(*,*) 'Two distinct circles found.'
    write(*,'(A10,F7.4,A1,F7.4)') 'Center1 : ', Center(1,1), ' ', Center(1,2)
    write(*,'(A10,F7.4,A1,F7.4)') 'Center2 : ', Center(2,1), ' ', Center(2,2)
  end if
elseif(Res.lt.0) then
  write(*,*) 'Incorrect value for r.'
end if
write(*,*)
end subroutine print_centers

subroutine test_inputs(P1,P2,R,Res)
implicit none
double precision :: P1(2), P2(2), R, dist
integer :: Res
if(R.lt.0.0d0) then
  Res = -1
  return
elseif(R.eq.0.0d0 .and. P1(1).eq.P2(1) .and. P1(2).eq.P2(2)) then
  Res = 1
  return
elseif(R.eq.0.0d0) then
  Res = 2
  return
elseif(P1(1).eq.P2(1) .and. P1(2).eq.P2(2)) then
  Res = 3
  return
else
  dist = sqrt( (P1(1)-P2(1))**2 + (P1(2)-P2(2))**2 )
  if(dist.gt.2.0d0*R) then
    Res = 4
    return
  else
    Res = 0
    return
  end if
end if
end subroutine test_inputs

subroutine find_center(P1,P2,R,Center)
implicit none
double precision :: P1(2), P2(2), MP(2), Center(2,2), R, dm, dd
MP = (P1 + P2)/2.0d0
dm = sqrt((P1(1) - P2(1))**2 + (P1(2) - P2(2))**2)
dd = sqrt(R**2 - (dm/2.0d0)**2)
Center(1,1) = MP(1) - dd*(P2(2) - P1(2))/dm
Center(1,2) = MP(2) + dd*(P2(1) - P1(1))/dm

Center(2,1) = MP(1) + dd*(P2(2) - P1(2))/dm
Center(2,2) = MP(2) - dd*(P2(1) - P1(1))/dm
end subroutine find_center