Conchoid of a circle

 

Here is the definition of conchoid of a curve.
Conchoid[4]. Let O be a fixed point and let L be a line through O intersecting a curve C at a point Q. The locus of points A and B on L such that AQ = QB = k, a constant, is a conchoid of C with respect to O.
We will construct using Jeometry the conchoid of a circle (assuming the constant k equal to its radius) and find the Cartesian equation using the Groebner applet.

Here is how the curve appears for a particular position of the fixed point O.

Let the circumference of radius r be centered at the origin origin (0, 0) and let O be (0, a), with a <> 0 and a <> r. A point Q(p, q) is on the circumference if and only if p2 + q2 = r2. The points A and B that trace out the conchoid are on the line OQ p(y - a) = x(q - a) and they satisfy the equation (x - p)2 + (y - q)2 = r2.
The reduced Groebner basis of I = <p2 + q2 - r2, p(y - a) - x(q - a), (x - p)2 + (y - q)2 - r2> w.r.t. lex order and p>q>x>y>r>a (if we use the Groebner applet we must treat r and a as variables) is

g1 = x6 + 3x4y2 - 2ax4y -4r2x4 + a2x4 + 3x2y4 - 4ax2y3 - 8r2x2y2
+ 2a2x2y2 + 8ar2x2y + y6 - 2ay5 - 4r2y4 + a2y4 + 8ar2y3 - 4a2r2y2
g2 = (a2y)q + 1/2x4 + x2y2 - 1/2ax4y - 2r2x2 + 1/2a2x2
+ 1/2y4 - 1/2ay3 - 2r2y2 - 1/2a2y2 + 2r2ay
g3 = (x2 + y2 - ay)q - 1/2x2y - 1/2ax2 - 1/2y3 + 1/2ay2
g4 = aq2 - 3/2qay - 1/4x2y + 1/4ax2 - 1/2y3 + 3/4ay2 + r2y - r2a
g5 = (y - a)p - qx + ax
g6 = px + qy - 1/2x2 - 1/2y2
g7 = qpa - 1/2pa2 - 3/2qax - 1/4x3 - 1/4xy2 + 1/2axy + r2x + 1/2a2x
g8 = p2 + q2 - r2

Hence the 2nd elimination ideal is given by

I2 = <g1> = <x6 + 3x4y2 - 2ax4y -4r2x4 + a2x4 + 3x2y4 - 4ax2y3 - 8r2x2y2
+ 2a2x2y2 + 8ar2x2y + y6 - 2ay5 - 4r2y4 + a2y4 + 8ar2y3 - 4a2r2y2>.

But does V(I2) introduce some extraneous points? This is the same as asking if every partial solution (x, y)extends to a complete solution (x, y, q, p). Over C we can give a positive answer using the Extension Theorem. And over R? Looking at the generators in the Groebner basis we see that we can extend the solution (x, y) to (x, y, q) (using g2, g3 and g4 when (x, y) = (0, 0)). Can we extend now (x, y, q) to (x, y, q, p)? If (x, y) <> (0, a)we can use g5 and g6 to see that p is real; otherwise, if (x, y) = (0, a) we have q = 1/2a from g7 = 0; substituting into g8 we get a real p if and only if |a| <= 2r. Hence, if |a| > 2r, V(I2)introduces one isolated point (0, a) which is not on the locus.

Now we will see what happens when a = r, i.e. the point O lays on the circumference. In such a case we have ambiguities in the equation p2 + q2 = r2 characterizing the point Q, since O too satisfies the same equation. This ambiguity can be easily removed by adding to I the new generator s p(q - r) - 1, which makes sure that O(0, r) is distinct from Q(p, q). Recomputing a Groebner basis for J = <p2 + q2 - r2, p(y - r) - x(q - r), (x - p)2 + (y - q)2 - r2, s p(q - r) - 1> w.r.t. lex order and s>p>q>x>y>r, we get

V(J3) = {(x, y) in R3: x4 + 2x2y2 - 3r2x2 + y4 - 3r2y2 + 2r3y = 0}.

In this case the curve is also known as the trisectrix.

Here are some screenshots of the conchoid for different positions of the fixed point O.

  1.  
    |a| > 2r. V(I2) covers the isolated point O.
  2.  
    |a| = 2r.
  3.  
    r < |a| < 2r.
  4.  
    |a| = r. The trisectrix.
  5.  
    0 < |a| < r.

 

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