Question

How do you convert `1=(-2x-2)^2+(y+7)^2` into polar form?

Answer 1

`r^2(4cos^2theta+sin^2theta)+r(14sintheta-8costheta)+52=0`

Explanation:

Given:
`1=(-2x-2)^2+(y+7)^2`
`x=rcostheta`
`y=rsintheta`
Now,
`1=(-2rcostheta-2)^2+(rsintheta+7)^2`
`1=4r^2cos^2theta-8rcostheta+4+r^2sin^2theta+14rsintheta+49`
`1=r^2(4cos^2theta+sin^2theta)+r(14sintheta-8costheta)+4+49`
Simplifying
`r^2(4cos^2theta+sin^2theta)+r(14sintheta-8costheta)+53-1=0`
`r^2(4cos^2theta+sin^2theta)+r(14sintheta-8costheta)+52=0`