Question 1131216
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I will disagree with tutors @MathTherapy and @ikleyn, both of whom say the solution is the empty set and the solution by tutor @MathLover1 is incorrect.<br>
In polar coordinates, r can be negative.  (-3,pi/2) is a valid description of a point in polar coordinates; so it is a point on the graph of r=-3.<br>
Any point with polar coordinates (3,theta) can be represented also by (-3,theta+pi); or with an infinite number of other representations with r equal to either 3 or -3.  So the graph of r=-3 is the same as the graph of r=3.<br>
If it were the case that r can't be negative, then simple polar equations like r=sin(theta) or r = -2+cos(theta) would be invalid.<br>
The equation r=-3 is independent of the angle theta; it is equivalent to the equation r=3.  Both equations are of a circle with center at the origin and radius 3.<br>
As tutor @MathLover1 said, the equation in rectangular coordinates is<br>
{{{x^2+y^2=9}}}