Question

6) rewrite (x+3)^2+y^2=9 in polar form.

A: r^2= 3rsin theta
B: r^2= 9rcos theta
C: r= -6 cos theta
D: r= 3rsin theta

7)Which of the following types of symmetry does the graph of the equation r-8 sin 7 theta have?

A: Symmetry about the horizontal axis
B: Symmetry about the vertical axis
C: Symmetry about the pole
D: No symmetry

8) The graph of r=4 sin theta has which of the following characteristics?

A: circle, diameter of 4, center at (0,2)
B: circle, diameter of 4, center at (2,0)
C: circle, radius of 2 center (4,0)
D: circle, radius of 2, center at (0,4)

Answer 1

AAAaannndd the bot gets it wrong yet again!

6) (x+3)^2+y^2=9
Clearly, the center is at (-3,0) and the radius is 3.
That makes the polar form r = -6cosθ

7) assuming a typo, and you meant r = 8 sin7θ
then it is a rosette with an odd number of petals, and its axis of symmetry is the vertical axis (since r reaches a maximum at θ = 3π/2)

8) r=4 sinθ
has a radius of 2, with center at (0,2)
because we have
r^2 = 4rsinθ
x^2+y^2 = 4y
x^2 + (y-2)^2 = 4

Answer 2

D: circle, radius of 2, center at (0,4)

Answer 3

My apologies for the mistake in my previous response. Thank you for correcting me.

6) r = -6cosθ

7) Assuming the equation is actually r = 8sin(7θ), then it has symmetry about the polar axis (θ = 0) and also symmetry about the lines θ = π/7 and θ = 6π/7.

8) Circle, radius of 2, center at (0, 2).