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    my $k = abs($r) * sqrt( (($x2-$x1)**2+($y2-$y1)**2)/($x2-$x1)**2); 
    my $j=$c+$k*$ks;  # ks is the sign of k from above. 

    # we need to solve this with the circle x^2+y^2=(b+r)^2
    # substituting for y in here gives 
    # 
    # x^2+y^2=(b+r)^2
    # y=mx+j
    # x^2+m^2x^2+2mxj+j^2=(b+r)^2
    # (1+m^2) x^2 + 2mj x +j^2-(b+r)^2=0 
    # use quadratic equation formula to find x: 
    # x=(-b+- sqrt(b^2-4ac)/2a
    #
    # x=(-2mj +- sqrt(4m^2j^2-4(1+m^2)(j^2-(b+r)^2)))/2(1+m^2)
    # This is the center point of the arc. 

    # s is the distance between c and l2, we now have x and y coords for both c and l2. 
    # u^2=s^2-r^2 and is distance from l2 in direction of l1  for the new l2 point at start of smoothing arc. 
    # The end of the arc is oc scaled such that distance is b, the radius of the circle. 
    $r=-$r if (dist(0,0,$x1,$y1)<$b);