Question

How do you convert `x^2+(y-4)^2=16` to polar form?

Answer 1

`r=8sintheta`

Explanation:

If `(r,theta)` is in polar form and `(x,y)` in Cartesian form the relation between them is as follows:

`x=rcostheta`, `y=rsintheta`, `r^2=x^2+y^2` and `tantheta=y/x`

Hence, `x^2+(y-4)^2=16` can be written as

`x^2+y^2-8y+16=16` or

`x^2+y^2-8y=0` or

`r^2-8rsintheta=0` or

`r(r-8sintheta)=0` dividing by `r`

`r-8sintheta=0` or

`r=8sintheta`

Answer 2

`r=8sintheta`

Explanation:

To convert from Cartesian to Polar coordinates use the following formulae that link them.

`• x=rcostheta" and " y=rsintheta`

`x^2+y^2-8y+16=16 " (expanding bracket) "`

`rArrr^2cos^2theta+r^2sin^2theta-8rsintheta+16-16=0`

then `r^2(cos^2theta+sin^2theta)-8rsintheta=0`

using the identity: `cos^2theta+sin^2theta=1`

`rArr r^2=8rsintheta" and dividing both sides by r "`

`rArr r=8sintheta`