# What is the equation of this curve?

1. Mar 14, 2010

### phantomcow2

1. The problem statement, all variables and given/known data

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 [Broken]

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.

Last edited by a moderator: May 4, 2017
2. Mar 14, 2010

### 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.

3. Mar 14, 2010

### 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!

Last edited: Mar 14, 2010
4. Mar 14, 2010

### 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)

5. Mar 14, 2010

### LCKurtz

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