Question

How do you convert `r=6 cos θ + 4 sin θ` into rectangular form?

Answer 1

`x^2 + y^2 - 6x - 4y = 0`

Explanation:

using the formulae that links Polar to Rectangular coordinates.

`• r^2 = x^2 + y^2`

`• x = rcostheta rArr costheta = x/r`

`• y = rsintheta rArr sintheta = y/r`

in the above question then

`r = 6.(x/r) + 4.(y/r)`

( multiplying both sides by r )

`r^2 = 6x + 4y`

`rArr x^2 + y^2 -6x - 4y = 0`

which is the equation of a circle : centre(3,2 ) and `r =sqrt13`