What is the equation of this curve?

[phantomcow2]

Homework Statement

Let C be the circle with center (a/2, 0) and radius a/2. Let L be the line with equation x=a, find the polar equation of the curve produced in the following manner: FOr every angle \theta, -\pi/2 <\theta < \pi/2, consider the ray from the origin that makes the angle \theta with the positive x axis. This ray will intersect C at a point A and L at a point B. Let O be the origin. Then the point P on the ray is on the curve if line segment OP = line segment AB.

http://img341.imageshack.us/img341/8910/cireq.jpg

If I trace it out, which I did in my example, it begins to look like a centroid. However I don't know how to verify this. I'm not sure exactly how to systematically approach this problem.
My idea is to break the equation of the resulting curve into two parts: first the length of the segment from the origin to when it hits the boundary of the circle, then the distance from the circle to the vertical line; the resulting curve being the difference between the former and the latter. I believe the distance of the line segment from O to A is simply cos\theta. I'm not sure how I'd express the rest of it though.

 

[LCKurtz]

Hints:

What is the equation of the line in polar coordinates: r₁ = ?? in terms of θ?

What is the equation of the circle in polar coordinates: r₂ = ?? in terms of θ?

Then your curve, in polar equations is r = r₁ - r₂.

The curve approaches x = a as a vertical asymptote.

 

[phantomcow2]

Darn, it's been too long since I've learned about polar coordinates. I know that the equation of the circle in polar coordinates is r=acos\theta. Some research is needed to find a straight line!

Thanks very much for your assistance!

 

[phantomcow2]

Okay, I looked through my old calculus books and have relearned that the equation of a vertical line is a=rcos(theta)

Resulting curve: acos(theta)-asec(theta)

 

[LCKurtz]

Yes, but you mean r = a sec(theta) - a cos(theta)