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

1. Jul 22, 2010

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution

First, I multiply both sides by r and distribute.

$$r^2=2hr\cos\theta+2kr\sin\theta$$

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

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

$$x^2+y^2=2(hx+ky)$$

not quite sure what do do from here.

The answer to the problem is supposed to be:

$$(x-h)^2+(y-k)^2=h^2+k^2; \sqrt{h^2+k^2}; (h,k)$$

2. Jul 22, 2010

### Staff: Mentor

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.

3. Jul 22, 2010

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

$$x^2+y^2=2hx+2ky$$

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

$$(x-h)^2+(y-k)^2=0$$

How would you get the $$h^2+k^2$$ on the RHS of the equation?

4. Jul 22, 2010

### Staff: Mentor

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. Jul 22, 2010

### jegues

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.

$$(x+2)^{2} = x^{2} + 4x + 4$$

or,

$$(x + 8)^{2} = x^{2} + 16x + 64$$

So now,

$$x^{2} - 2hx + ?$$

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