algebra-precalculuscirclespolar coordinates
I am trying to graph a circle using a polar equation that does not intersect the pole. For example the rectangular equation would be $(x-3)^2+(y+2)^2=1$
If I use the usual way to transform this into a polar equation I end up with the equation $r^2=-12+6\cos\theta-4\sin\theta$, and if I take the square root of both sides I get r equal to an imaginary number which can't be graphed in the real plane. Any suggestions for another way to do this?
The two square roots of a complex $w$ with module $\left\vert w\right\vert $ and $\theta =\arg (w)$, i.e $w=\left\vert w\right\vert e^{i\theta }$ are $$ \sqrt{w}=\sqrt{\left\vert w\right\vert }e^{i\frac{\theta +2k\pi }{2}},k\in \{0,1\}.$$
For $w=i$, we have $\left\vert w\right\vert =\left\vert i\right\vert =1$, $% \theta =\arg (i)=\frac{\pi }{2}$. Thus $$\sqrt{e^{i\pi /2}}=e^{i\frac{\pi /2+2k\pi }{2}},k\in \{0,1\}.$$ One of the roots is
$$z_{1}=e^{i\frac{\pi /2}{2}}=e^{i\pi /4}=\cos \left( \frac{\pi }{4}\right) +i\sin \left( \frac{\pi }{4}\right) =\frac{1}{2}\sqrt{2}+\frac{1}{2}i\sqrt{2} $$ and the other $$z_{2}=e^{i\frac{\pi /2+2\pi }{2}}=e^{i5\pi /4}=\cos \left( \frac{5}{4% }\pi \right) +i\sin \left( \frac{5}{4}\pi \right) =-\frac{1}{2}\sqrt{2}-% \frac{1}{2}i\sqrt{2}.$$
Only $z_{2}$ has negative imaginary part. Observing that $\frac{1}{2}\sqrt{2}% =\frac{1}{\sqrt{2}}$, we get $$z_{2}=\frac{1}{\sqrt{2}}\left( -1-i\right) .$$
Added 2: $w=i,z_1,z_2$
Added: The $n^{th}$ roots of the complex $w=\left\vert w\right\vert e^{i\theta }$ are
$$\sqrt[n]{w}=\sqrt[n]{\left\vert w\right\vert }e^{i\frac{\theta +2k\pi }{n}},k\in \left\{ 0,1,2,\ldots ,n-1\right\} ,$$
because
$$\begin{eqnarray*} \left( \sqrt[n]{w}\right) ^{n} &=&\left( \sqrt[n]{\left\vert w\right\vert }% e^{i\frac{\theta +2k\pi }{n}}\right) ^{n}=\left( \sqrt[n]{\left\vert w\right\vert }\right) ^{n}\left( e^{i\frac{\theta +2k\pi }{n}}\right) ^{n} \\ &=&\left\vert w\right\vert e^{i\left( \theta +2k\pi \right) }=\left\vert w\right\vert e^{i\theta }e^{i2k\pi }=\left\vert w\right\vert e^{i\theta }\cdot 1 \\ &=&\left\vert w\right\vert e^{i\theta }=w. \end{eqnarray*}$$
A circle is described in polar form by $r^2-2rx_0\cos(\theta - y_0)+x_0^2=a^2$ where $a$ is a constant and $x_0$ and $y_0$ represent the coordinates of the center of the circle. Therefore, $$\frac{dr}{d\theta}=2rx_0\sin(\theta-y_0)$$ So if the equation you are given describes a circle, then $\frac{dr}{d\theta}=2rx_0\sin(\theta-y_0)$. All you have to do is prove this. Implicit differentiation will be needed, though.
Best Answer
I assume that you replaced $x$ with $r \cos \theta$ and $y$ with $r \sin \theta$ in the equation $(x-3)^2 + (y+2)^2 = 1$ to get $r^2 = -12 + 6 \cos \theta - 4 \sin \theta$. Since $\cos \theta$ and $\sin \theta$ take values between -1 and 1, that implicitly assumes that the circle is centered around the origin, whereas your circle is centered around (3,-2). The correct set of polar equations is $$ \begin{cases} x = \cos \theta + 3 \\ y = \sin \theta - 2 \end{cases} $$
In general, for an equation $(x-x_c)^2 + (y-y_c)^2 = r^2$, the corresponding set of polar equations is $$ \begin{cases} x = r \cos \theta + x_c \\ y = r \sin \theta + y_c \end{cases} $$