Sol A gives the polar equation , so taking square roots gives . Therefore is a polar equation for the circle (since and represent the same circle).
Sol B Multiplying both sides of by gives , so . To put this equation in standard form, we can subtract from both sides to get and then complete the square to obtain or .
Sol C Taking gives , so is one set of polar coordinates for the point. Taking gives , so is another set of polar coordinates for the point.
(More generally, we can take and to be any odd multiple of , or and to be any even multiple of .
Sol 1 , so we can take . Since and is in Quadrant IV, we can take . Therefore is a set of polar coordinates for the point.
Sol 2 , so we can take . Since and is in Quadrant III, we can take . Therefore is a set of polar coordinates for the point.
Sol 3 and , so the point has rectangular coordinates .
Sol 4 gives , so and therefore .
Sol 5 gives , so and therefore .
Sol 6 gives or , so and therefore .
Sol 7 Multiplying both sides of by gives or . Then or , so squaring both sides gives , and thus .
Sol 8 Setting the two expressions for equal to each other gives , so and . Therefore we can take or , and then the corresponding value of is given by . Therefore the curves intersect at the points with polar coordinates and , and they also intersect at the origin by inspection.
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