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

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Answer 1 · 2016-10-24T22:06:06

Please see the 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.

How do you convert 9=(x-4)^2+(y+2)^29=(x-4)^2+(y+2)^2 into polar form?