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A) In order to convert that rectangular coordinates into a polar one, we need to think of a right triangle whose hypotenuse is connecting the point to the origin.
So, we need to resort to some equations:
[tex]\begin{gathered} x^2+y^2=r^2 \\ \tan(\theta)=\frac{y}{x} \\ \theta=\tan^{-1}(\frac{y}{x}) \end{gathered}[/tex]
Thus, we need now to plug x=-4 and y=4 into that:
[tex]\begin{gathered} r=\sqrt{\left(-4\right)^2+4^2}\Rightarrow r=4\sqrt{2} \\ θ=\arctan \left(\frac{4}{-4}\right) \\ θ=\arctan\left(\frac{4}{-4}\right)+\pi \\ θ=-\frac{\pi }{4}+\pi \end{gathered}[/tex]
Note that we needed to add pi to the arctangent to adjust that point to the Quadrant.
Thus, the answer is:
[tex]\left(4\sqrt{2},\:-\frac{\pi }{4}+\pi \right)[/tex]
B) What is the polar form of the equation? What type of polar curve is this?
Let's convert that to the polar form:
[tex]\begin{gathered} x=r\cos^(\theta) \\ y=r\sin(\theta) \\ (r(\cos(\theta))^2+(r(\sin(\theta))^2+8(r\cos(\theta)=0 \\ r^2\cos^2(\theta)+r^2\sin^2(\theta)+8r\cos(\theta)=0 \\ r\left(r\cos^2\left(θ\right)+r\sin^2\left(θ\right)+8\cos\left(θ\right)\right)=0 \\ r(r(\cos^2\left(θ\right)+\sin^2\left(θ\right))+8\cos(\theta))=0 \\ r(r(1)+8cos(\theta))=0 \\ \frac{r(r(1)+8cos(\theta))}{r}=\frac{0}{r} \\ r+8\cos(\theta)=0 \\ r=-8\cos(\theta) \end{gathered}[/tex]
Note that we had to simplify that because Polar equations are linear, we have also used a trigonometry identity. Since this polar equation is in the form:
[tex]r=\pm a\cos(\theta)[/tex]
Then this type of Polar Curve is a Circle.
