subroutine circen(i,j,k,x0,y0,x,y,ntot,eps,collin,nerror)
# Find the circumcentre (x0,y0) of the triangle with
# vertices (x(i),y(i)), (x(j),y(j)), (x(k),y(k)).
# Called by qtest1, dirseg, dirout.

implicit double precision(a-h,o-z)
dimension x(-3:ntot), y(-3:ntot), xt(3), yt(3)
logical collin

nerror = -1

# Get the coordinates.
xt(1) = x(i)
yt(1) = y(i)
xt(2) = x(j)
yt(2) = y(j)
xt(3) = x(k)
yt(3) = y(k)

# Check for collinearity
ijk = 0
call cross(xt,yt,ijk,cprd)
if(abs(cprd) < eps) collin = .true.
else collin = .false.

# Form the vector u from i to j, and the vector v from i to k,
# and normalize them.
a  = x(j) - x(i)
b  = y(j) - y(i)
c  = x(k) - x(i)
d  = y(k) - y(i)
c1 = sqrt(a*a+b*b)
c2 = sqrt(c*c+d*d)
a  = a/c1
b  = b/c1
c  = c/c2
d  = d/c2

# If the points are collinear, make sure that they're in the right
# order --- i between j and k.
if(collin) {
        alpha = a*c+b*d
        # If they're not in the right order, bring things to
        # a shuddering halt.
        if(alpha>0) {
		nerror = 3
		return
        }
        # Collinear, but in the right order; think of this as meaning
        # that the circumcircle in question has infinite radius.
        return
}

# Not collinear; go ahead, make my circumcentre.  (First, form
# the cross product of the ***unit*** vectors, instead of the
# ``normalized'' cross product produced by ``cross''.)
crss = a*d - b*c
x0 = x(i) + 0.5*(c1*d - c2*b)/crss
y0 = y(i) + 0.5*(c2*a - c1*c)/crss

return
end