Finding polar equation of a shifted cricle

[farfromdaijoubu]

Homework Statement

The question asked to select the correct polar equation of a circle of radius 4 that had been shifted up one unit, so x²+(y-1)²=16.

Relevant Equations

x²+(y-1)²=16
x=rcos(θ) and y=rsin(θ

The question asked to select the correct polar equation of a circle of radius 4 that had been shifted up one unit, so x²+(y-1)²=16.

I subbed in x=rcos(θ) and y=rsin(θ) on the left-hand side and after some algebra obtained the quadratic r² -2rsin(θ) - 15 = 0. Solving this, I got r=sin(θ) ± √(sin²(θ)+15).

But I don't know how to proceed from here to get the answer r=4+sin(θ)

 

[Steve4Physics]

The official answer, r=4+sin(θ), is wrong. Look at this link.

 

[farfromdaijoubu]

Ahh thank you. It was my mistake then for thinking this given image was a circle..

 

[Steve4Physics]

Ahh thank you. It was my mistake then for thinking this given image was a circle..View attachment 361469

The height on your diagram is 8. But - check the scale marks - the width is more than 8. So it's not circular.

For information, a curve of the form $r = a + b \sin \theta$ (or $r = a + b \cos \theta$) is called a ‘limaçon’.

 

[Mark44]

At first glance, one could think that the image was that of a circle. For a circle centered at (0, 0), of radius 4, and translated up by 1 unit, the x-intercepts would be at $x = \pm \sqrt{15} \approx \pm 3.87$.

What information was given in the problem? Just based on the image, it would be a stretch to come up with the Cartesian equation, let alone the polar equation.

 

[farfromdaijoubu]

At first glance, one could think that the image was that of a circle. For a circle centered at (0, 0), of radius 4, and translated up by 1 unit, the x-intercepts would be at $x = \pm \sqrt{15} \approx \pm 3.87$.

What information was given in the problem? Just based on the image, it would be a stretch to come up with the Cartesian equation, let alone the polar equation.

It was a multiple choice with just the image but all the options were of the form Steve4Physics said. At the time I just ended up just plotting each one and figuring it out that way but I should've recognized it was a limacon.

 

[Gavran]

Also, the curve looks like an ellipse. $$ \frac{15x^2}{256}+\frac{(y-1)^2}{16}=1 $$ If all the options are of the form $r=a+b\sin\theta$ or $r=a+b\cos\theta$ where $a\neq0$ and $b\neq0$, it is definitely a limacon.