Question

the polar equation r=6cosθ - 2sinθ is the polar form of a circle. determine the center and radius of this circle algebraically

Answer 1

Have you considered multiplying by "r"?

Answer 2

No I haven't. what will that do?

Answer 3

It's close to magic. Give it a go.

Answer 4

Would you get \[r ^{2}=6\cos \Theta r - 2 \sin \Theta r?\] Is there more you can do?

Answer 5

It might help if you would write it this way:

\(r^{2} = 6(r\cos(\theta)) - 2(r\sin(\theta))\)

Answer 6

Okay. When I look at it this way do the 6 and 2 happen to be the center? R^2 I know in normal circles equals the radius squared does this have significance here

Answer 7

So close!!! Not quite.

Change it to rectangular coordinates and check out your theory.

Answer 8

Dang! How do I change them to regular coordinates?

Answer 9

\(r^{2} = x^{2} + y^{2}\)

\(x = r\cos(\theta)\)

\(y = r\sin(\theta)\)

Answer 10

So would it be \[6=r \cos \Theta and -2=r \sin \theta \]

Answer 11

?? Something fell apart, there

\(r^{2} = 6r\cos(\theta) - 2r\sin(\theta) \rightarrow x^{2} + y^{2} = 6x - 2y\)

Give that a good hard look until you see it.

Answer 12

Hmm.. Is that a formula? It isn't ringing a bell? :( I know it should be something

Answer 13

It was ONLY the three substitutions I gave you.

\(r^{2} = 6r\cos(\theta) - 2r\sin(\theta)\)

Given \(r^{2} = x^{2} + y^{2}\), we have

\(x^{2} + y^{2} = 6r\cos(\theta) - 2r\sin(\theta)\)

Given \(x = r\cos(\theta)\), we have

\(x^{2} + y^{2} = 6x - 2r\sin(\theta)\)

Given \(y = r\sin(\theta)\), we have

\(x^{2} + y^{2} = 6x - 2y\)

Substitution. That's all it is.

Answer 14

Oh okay. I see it now with the break down. Thank you. How does the center and radius come out of that. I think I am seeing how to get the center but not the radius

Answer 15

Time for your best "Completing the Square" skills. How's your algebra? Now's the time to prove it!

Answer 16

I think they are pretty good. I will work it out and then show what I got

Answer 17

Okay. So I got\[x ^{2}- 6x +9 or (x-3)^{2}. \]. For Y I got\[-y ^{2}-2y+1\]which isn't a perfect square. Does the y remain positive?

Answer 18

You DO have to do them together.

\(x^{2} + y^{2} = 6x - 2y\)

Set it up

\(x^{2} - 6x + y^{2} + 2y = 0\)

Complete the Square on x

\((x^{2} - 6x + 9) + y^{2} + 2y = 9\)

Complete the Square on y

\((x^{2} - 6x + 9) + (y^{2} + 2y + 1) = 9 + 1\)

Rewrite

\((x-3)^{2} + (y+1)^{2} = 10\)

You seem to have the idea. You just have to be a little more organized.

Answer 19

Okay. Perfect. So this should mean looking at the equation that the center is is at (3,-1). Would the radius be 10 or root 10? The equation looks so familiar.

Answer 20

Rewrite: \((x-3)^{2} + (y+1)^{2} = \left(\sqrt{10}\right)^{2}\)

What we need to learn is what the original equation says. This will save us from converting the world to Rectangular Coordinates.

\(r=6cosθ - 2sinθ\)

See how that +6 turns into -3?
See how that -2 turns into +1?
See how 3 and 1 turn into 10? \(3^{2} + 1^{2}\)

These are the lessons we must learn.

Answer 21

Aww okay. It is beginning to make sense. They are squaring those numbers. I can start to see it. Math is so tricky yet cool.