Converting A Polar Equation to Rectangular Form; Equation of a Circle

  • #1
themadhatter1
140
0

Homework Statement


Convert the polar equation

r = 2(h cos θ + k sin θ)

to rectangular form and verify that it is the equation of a circle. Find the radius and the rectangular coordinates of the center of the circle.

Homework Equations





The Attempt at a Solution



First, I multiply both sides by r and distribute.

[tex]r^2=2hr\cos\theta+2kr\sin\theta[/tex]

apply the x= r cos θ and y= r sin θ equations

[tex]r^2=2hx+2ky[/tex]
from here I can factor out the 2 and plug it into the equation for a circle.

[tex]x^2+y^2=2(hx+ky)[/tex]

not quite sure what do do from here.

The answer to the problem is supposed to be:

[tex](x-h)^2+(y-k)^2=h^2+k^2; \sqrt{h^2+k^2}; (h,k)[/tex]
 

Answers and Replies

  • #2
36,311
8,281

Homework Statement


Convert the polar equation

r = 2(h cos θ + k sin θ)

to rectangular form and verify that it is the equation of a circle. Find the radius and the rectangular coordinates of the center of the circle.

Homework Equations





The Attempt at a Solution



First, I multiply both sides by r and distribute.

[tex]r^2=2hr\cos\theta+2kr\sin\theta[/tex]

apply the x= r cos θ and y= r sin θ equations

[tex]r^2=2hx+2ky[/tex]
from here I can factor out the 2 and plug it into the equation for a circle.

[tex]x^2+y^2=2(hx+ky)[/tex]

not quite sure what do do from here.
So far, so good. Separate the terms on the right, and bring them over to the left. Then complete the squares in the x and y terms.
The answer to the problem is supposed to be:

[tex](x-h)^2+(y-k)^2=h^2+k^2; \sqrt{h^2+k^2}; (h,k)[/tex]
 
  • #3
themadhatter1
140
0
So far, so good. Separate the terms on the right, and bring them over to the left. Then complete the squares in the x and y terms.

Hmm. Ok, I think I know what you mean.

[tex]
x^2+y^2=2hx+2ky
[/tex]

bring it over to the other side and complete the square and you get

[tex](x-h)^2+(y-k)^2=0[/tex]

How would you get the [tex]h^2+k^2[/tex] on the RHS of the equation?
 
  • #4
36,311
8,281
Hmm. Ok, I think I know what you mean.

[tex]
x^2+y^2=2hx+2ky
[/tex]

bring it over to the other side and complete the square and you get

[tex](x-h)^2+(y-k)^2=0[/tex]

How would you get the [tex]h^2+k^2[/tex] on the RHS of the equation?

You're skipping the steps that would produce what you're looking for.
x2+y2=2hx+2ky
==> x2 - 2hx +y2 - 2ky = 0

Now, when you complete the squares in the x and y terms what do you need to add? You'll need to add the same amount on the right side.
 
  • #5
jegues
1,097
3
Now, when you complete the squares in the x and y terms what do you need to add?

If it still isn't obvious what you need to add on each side of your equation, simply expand some squares to gain some insight.

What about,

[tex] (x+2)^{2} = x^{2} + 4x + 4 [/tex]

or,

[tex] (x + 8)^{2} = x^{2} + 16x + 64 [/tex]

So now,

[tex] x^{2} - 2hx + ? [/tex]

What do you need to add (maybe in terms of h :wink:) in order to complete the square?
 

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