Question

How do you convert `9=(x-4)^2+(y+2)^2` into polar form?

Answer 1

Please see the explanation

Explanation:

Multiply the squares:

`9 = x^2 - 8x + 16 + y^2 + 4y + 4`

Combine like terms:

`0 = x^2 + y^2 - 8x + 4y + 11`

Substitute `r^2` for `x^2 + y^2`, `rcos(theta)` for x, and `rsin(theta)` for y:

`0 = r^2 - 8rcos(theta) + 4rsin(theta) + 11`

Group a common factor of 4r from the middle terms:

`0 = r^2 + 4(sin(theta) - 2cos(theta))r+ 11`

This is a quadratic in r, therefore, use the quadratic formula:

`r = {2(2cos(theta) - sin(theta)) +- sqrt(16(sin(theta) - 2cos(theta))^2 - 4(11))}/2`

Remove a factor of 4 from under the square root and make it 2 outside the square root:

`r = {2(2cos(theta) - sin(theta)) +- 2sqrt(4(sin(theta) - 2cos(theta))^2 - 11)}/2`

`2/2` becomes 1:

`r = 2cos(theta) - sin(theta) +- sqrt(4(sin(theta) - 2cos(theta))^2 - 11)`

If wish, you can drop the negative square root without losing any an points on the circle.