r = -3, squaring both sides we get r^{2} = 9 then r^{2}(cos^{2} q + sin^{2} q) = 9

\ r^{2} cos^{2} q + r^{2} sin^{2} q = 9

\ (r cos q)2 + (r sin q)2 = 9

put r cos q = x and r sin q = y we get x^{2} + y^{2} = 9 which is a circle of radius '3' and centre (0, 0).

r = a cos q

\ r^{2} = ra cos
q

\ r^{2}(cos^{2} q + sin2 q) = a (r cos q)

\ r^{2} cos^{2} q + r^{2} sin^{2} q = a (r cos q)

\ (r cos q)^{2} + (r sin q)^{2} = a (r cos q)

\ (x)^{2} + (y)^{2} = a (x) ... x = r cos
q and y = r sin q

\ x^{2} + y^{2} - ax = 0

r = b sin q \ r^{2} = b r sin q

\ x^{2} + y^{2} = b (y)

since r^{2} = x^{2} + y^{2} and r sin q = y

\ x^{2} + y^{2} - by = 0

q = a which is a straight line through origin.

r = b cosec q

\ r =

\ r sin q = b

\ y = b which is a straight line parallel to x - axis at a distance 'b' unit from the origin.