Question

How do you convert `4=(x-2)^2+(y-6)^2` into polar form?

Answer 1

`r=2costheta+6sintheta+-2sqrt(3sin2theta-8cos^2theta)`

Explanation:

Use the rule for switching from a Cartesian system to a polar system:

• `x=rcostheta`
• `y=rsintheta`

This gives:

`4=(rcostheta-2)^2+(rsintheta-6)^2`

`4=r^2cos^2theta-4rcostheta+4+r^2sin^2theta-12rsintheta+36`

`0=r^2cos^2theta+r^2sin^2theta-4rcostheta-12rsintheta+36`

`0=r^2(cos^2theta+sin^2theta)+r(-4costheta-12sintheta)+36`

Note that `cos^2theta+sin^2theta=1` through the Pythagorean identity.

`0=r^2+r(-4costheta-12sintheta)+36`

This can be solved through the quadratic equation.

`r=(4costheta+12sintheta+-sqrt(16cos^2theta+96costhetasintheta+144sin^2theta-144))/2`

`r=(4costheta+12sintheta+-4sqrt(cos^2theta+6costhetasintheta+9sin^2theta-9))/2`

`r=2costheta+6sintheta+-2sqrt(cos^2theta+6costhetasintheta+9sin^2theta-9)`

`r=2costheta+6sintheta+-2sqrt(cos^2theta+6costhetasintheta+9(sin^2theta-1))`

`r=2costheta+6sintheta+-2sqrt(cos^2theta+6costhetasintheta-9cos^2theta)`

`r=2costheta+6sintheta+-2sqrt(6costhetasintheta-8cos^2theta)`

Note that `2costhetasintheta=sin2theta`, so `6costhetasintheta=3sin2theta`.

`r=2costheta+6sintheta+-2sqrt(3sin2theta-8cos^2theta)`

If you have any questions as to any of the underlying algebra going on here, feel free to ask. I felt it would be a little much to try to explain each step.