Question

Convert the polar equation r= 6cos θ - 8sin θ to rectangular equation?

Answer 1

See explanation below

Explanation:

In polar coordinates we know that

`x=rcostheta`
`y=rsintheta`

From here we express `costheta=x/r` and `sintheta=y/r`

In our equation

`r=6costheta-8sintheta=6x/r-8y/r` multiplying by r we have

`r^2=6x-8y` but by definition `x^2+y^2=r^2` and then

`x^2+y^2=6x-8y` completing squares

`(x-3)^2+(y+4)^2-9-16=(x-3)^2+(y+4)^2-25=0`

The equation represents a circle with center `(3,-4)` and radious 5