Polar Coordinates

Sol A $x^2+y^2=25$ gives the polar equation $r^2=25$, so taking square roots gives $r=\pm5$. Therefore $r=5$ is a polar equation for the circle (since $r=5$ and $r=-5$ represent the same circle).

Sol B Multiplying both sides of $r=4\cos\theta$ by $r$ gives $r^2=4r\cos\theta$, so $x^2+y^2=4x$. To put this equation in standard form, we can subtract $4x$ from both sides to get $x^2-4x+y^2=0$ and then complete the square to obtain $(x^2-4x+4)+y^2=4$ or $(x-2)^2+y^2=4$.

Sol C Taking $\theta=\pi$ gives $r=3$, so $(3,\pi)$ is one set of polar coordinates for the point. Taking $\theta=0$ gives $r=-3$, so $(-3,0)$ is another set of polar coordinates for the point.

(More generally, we can take $r=3$ and $\theta$ to be any odd multiple of $\pi$, or $r=-3$ and $\theta$ to be any even multiple of $\pi$.

Sol 1 $r^2=x^2+y^2=(4\sqrt{3})^2+(-4)^2=48+16=64$, so we can take $r=8$. Since $\tan\theta=y/x=(4\sqrt{3})/(-4)=-\sqrt{3}$ and $\theta$ is in Quadrant IV, we can take $\theta=2\pi-\pi/3=5\pi/3$. Therefore $(8,5\pi/3)$ is a set of polar coordinates for the point.

Sol 2 $r^2=x^2+y^2=(-5)^2+(-5)^2=25+25=50$, so we can take $r=\sqrt{50}=5\sqrt{2}$. Since $\tan\theta=y/x=(-5)/(-5)=1$ and $\theta$ is in Quadrant III, we can take $\theta=5\pi/4$. Therefore $(5\sqrt{2},5\pi/4)$ is a set of polar coordinates for the point.

Sol 3 $x=r\cos\theta=12\cos 4\pi/3=12(-1/2)=-6$ and $y=r\sin\theta=12\sin 4\pi/3=12(-\sqrt{3}/2)=-6\sqrt{3}$, so the point has rectangular coordinates $(-6,-6\sqrt{3})$.

Sol 4 $x=4$ gives $r\cos\theta=4$, so $r=\frac{4}{\cos\theta}$ and therefore $r=4\sec\theta$.

Sol 5 $y=6$ gives $r\sin\theta=6$, so $r=\frac{6}{\sin\theta}$ and therefore $r=6\csc\theta$.

Sol 6 $(x-3)^2+y^2=9$ gives $x^2-6x+9+y^2=9$ or $x^2+y^2=6x$, so $r^2=6r\cos\theta$ and therefore $r=6\cos\theta$.

Sol 7 Multiplying both sides of $r=5(1-\sin\theta)$ by $r$ gives $r^2=5(r-r\sin\theta)$ or $x^2+y^2=5(r-y)$. Then $x^2+y^2=5r-5y$ or $x^2+y^2+5y=5r$, so squaring both sides gives $(x^2+y^2+5y)^2=25r^2$, and thus $(x^2+y^2+5y)^2=25(x^2+y^2)$.

Sol 8 Setting the two expressions for $r$ equal to each other gives $6\cos\theta=2+2\cos\theta$, so $4\cos\theta=2$ and $\cos\theta=1/2$. Therefore we can take $\theta=\pi/3$ or $\theta=5\pi/3$, and then the corresponding value of $r$ is given by $r=6\cos\theta=6(1/2)=3$. Therefore the curves intersect at the points with polar coordinates $(3,\pi/3)$ and $(3,5\pi/3)$, and they also intersect at the origin $(0,0)$ by inspection.

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