Why is the graph of $r = a + b\cos \theta$ the same whether a is positive or negative?

Question

So today's lecture was about polar coordinates, and we were taught about the concept up to limacons. I'd like to know why the graph of $r = a + b\cos \theta$ is exactly the same as the graph of $r = -a + b\cos \theta$ . I've tried substituting values for a and b but I still can't make sense of the results.

Answer 1

$r=a+b \cos \theta$ and $r=-a+b \cos \theta$ look very much the same, but are offset by $\pi$ and negated.

In a normal plot, here with a=5,b=7, plotted from $\theta=-\pi .. \pi$ you see the difference:

+a:

-a:

If you plot the whole graph on polar plot, the graph is cyclic, so it does not matter where you start or end. The offset by $\pi$ rotates the plot by 180° and the negation of the radius rotates it again by 180°. So both effects chancel each other and the polar graphs look exactly the same. If you only plot a part of the graph you see the difference.

Here the full Polar Graph of both cases (a=5, b=7):

Now the +a case only plotted from $\theta = -\pi .. \pi/2$:

and the -a case only plotted from $\theta = -\pi .. \pi/2$:

Answer 2

When you plot the cardoid in polar coordinates it settles all your doubts.

Assuming $\theta=0$ corresponds to x-axis counterclockwise rotation:

For $ a > 0, b > 0 $

$ a > b, \theta =0 $ gives two extreme cutting points on the cardoid.

$ a < b, \theta =0 $ gives four extreme cutting points on the cardoid.

$ a = 0 $ is a circle of diameter $b$ tangential to y-axis due to three coincident points.

Now you see $ \cos \theta $ can be $ >0$ or $<0$ for $ \theta = 0 , \pi $

So the plot makes two negatives multiplicatively positive at some places, and negative radius vector can have the tip of it in any of the 4 quadrants.