Calculus: Early Transcendentals (2nd Edition)

$(x - 3)^2 + (y + 1)^2 = 10$ This is a circle.
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.