Answer
$(x - 3)^2 + (y + 1)^2 = 10$
This is a circle.
Work Step by Step
Recall:
$r sin θ = y$
$r cos θ = x$
$r^2 = x^2 + y^2$
A circle has the equation $(x - h)^2 + (y - k)^2 = r^2$
Thus:
Problem: $r^2 + r (2 sin θ - 6 cos θ) = 0$
Distribute $r$: $r^2 + 2r sin θ - 6r cos θ = 0$
Convert from polar to rectangular: $x^2 + y^2 + 2y - 6x = 0$
Reorder equation: $x^2 - 6x + y^2 + 2y = 0$
Complete the square: $x^2 - 6x + 9 + y^2 + 2y + 1 = 0 + 9 + 1$
Simplify: $(x - 3)^2 + (y+1)^2 = 10$
This is a circle.
