Question

How do I convert (x-4)^2+(y-3)^2=25 to a polar equation?

Answer 1

Use mathway.com

Answer 2

did you use the website it really helps me with my math question

Answer 3

Trust me, I did. They're good for everything except this. It really pisses me off.

Answer 4

First thing you should do is expand the equation. Have you try it?

Answer 5

Once you get this, you should try the change:
\[x=r\cos\theta \\ y=r\sin\theta\]

Answer 6

I have, but I only made it as far as \[r^2\cos^2\theta+8rcos \theta+25+r^2\sin^2\theta+6r \sin \theta=25\]

Answer 7

Remember that, in this case, r=5 and that:
\[r^2\cos^2\theta+r^2\sin^2\theta=r^2\]

Answer 8

You should simplify what you have correctly obtained.

Answer 9

\[r^2(\cos^2\theta+\sin^2\theta)=r^2\]

Answer 10

Yes

Answer 11

What is the final result do you get then?

Answer 12

r^2, I suppose.

Answer 13

No, this is a property you are using to simplify this result you gave me

\[r^2\cos^2θ+8r\cosθ+25+r^2\sin^2θ+6r\sinθ=25\]

Answer 14

If you simplify this, you will get this,
\[r^2-8r\cos\theta-6r\sin\theta=0\]Ok?

Answer 15

So this would mean \[r^2-(8rcos \theta - 6rsin \theta)=0\]?

Answer 16

Sorry, your result was wrong with the signs, the correct one is:

\[r^2\cos^2θ-8r\cosθ+25+r^2\sin^2θ-6r\sinθ=25\]

Answer 17

Then you get the one I gave,

\[r^2−8r\cosθ−6r\sinθ=0\]

Answer 18

Now you can do the following, extract common factor r,
\[r(r-8\cos\theta-6\sin\theta)=0\]

Answer 19

So you have two options,
\[r=0\] And
\[r-8\cos\theta-6\sin\theta=0\] You need the last one, that can be rewritten as,
\[r=8\cos\theta+6\sin\theta\] And this is the polar equation.

Answer 20

Do you understand it?

Answer 21

You can check the graph here also,
https://www.wolframalpha.com/input/?i=PolarPlot%5B8+cos%5Bx%5D%2B6+sin%5Bx%5D,%7Bx,0,5%7D%5D

Answer 22

Thanks a million.

Answer 23

You're welcome.