# Polar to rectangular form.

• Jul 30th 2008, 05:32 AM
dashreeve
Polar to rectangular form.
I'm assuming this is a circle and will end up in x^2+y^2=r^2.

r=-8cos0 (0 being Omega)
Should I start by squaring each side knowing that r^2 will be in the rectangular form?? A little lost, not sure where to begin.
• Jul 30th 2008, 06:05 AM
Chris L T521
Quote:

Originally Posted by dashreeve
I'm assuming this is a circle and will end up in x^2+y^2=r^2.

r=-8cos0 (0 being Omega)
Should I start by squaring each side knowing that r^2 will be in the rectangular form?? A little lost, not sure where to begin.

The thing that you need to do here is multiply both sides by r:

$\displaystyle r(r)=-8(r)\cos\varpi$

$\displaystyle \implies r^2=-8r\cos\varpi$

Thus, in rectangular form, we get $\displaystyle x^2+y^2=-8x$

$\displaystyle \implies x^2+8x+y^2=0\implies x^2+8x+16+y^2=16\implies \color{red}\boxed{(x+4)^2+y^2=16}$

Does this make sense?

--Chris
• Jul 30th 2008, 06:12 AM
dashreeve
That's where I was going wrong, by squaring each side. When you put the 16 in, is that like completing the square?
• Jul 30th 2008, 06:19 AM
Chris L T521
Quote:

Originally Posted by dashreeve
That's where I was going wrong, by squaring each side. When you put the 16 in, is that like completing the square?

Yes, it appears as a result of completing the square. Does this seem to make more sense now? This polar stuff can get a little confusing.

--Chris
• Jul 30th 2008, 06:26 AM
dashreeve
Yes it makes much more since now, thanks for the help.