Question

Convert `(x+2)^2 + (y+3)^2 = 13` into polar form?

Answer 1

`r =- 4costheta - 6sintheta`

Explanation:

Putting:

`x = rcos theta`
`y = rsin theta`

Into the cartesian equation:

`(x+2)^2 + (y+3)^2 = 13`

we have:

`(rcos theta+2)^2 + (rsin theta+3)^2 = 13`

`:. r^2cos^2 theta+4rcostheta + 4 + r^2sin^2theta+6rsintheta+9 = 13`

`:. r^2(cos^2 theta + sin^2theta) + 4rcostheta + 6rsintheta+13 = 13`

`:. r^2 + 4rcostheta + 6rsintheta = 0`

`:. r(r + 4costheta + 6sintheta) = 0`

Leading to two possibilities:

`{ (r = 0), (r =- 4costheta - 6sintheta) :}`

Note that the second equation encompases the first when:

`6sintheta = 4costheta => tan theta = 2/3`

Hence the Polar equation we seek is:

`r =- 4costheta - 6sintheta`