Question

How do you convert `2sinθ-3cosθ=r` to rectangular form?

Answer 1

The answer is `(x+3/2)^2+(y-1)^2=13/4`

Explanation:

To convert from polar coordinates `(r,theta)` to rectangular coordinates `(x,y)`, we use the following equations

`x=rcostheta`, `=>`, `costheta=x/r`

`y=rsintheta`, `=>`, `sintheta=y/r`

`x^2+y^2=r^2`

Therefore

`2sintheta-3costheta=r`

`2/ry-3/rx=r`

`2y-3x=r^2`

`2y-3x=x^2+y^2`

`x^2+3x+y^2-2y=0`

`(x^2+3x+9/4)+(y^2-2y+1)=9/4+1`

`(x+3/2)^2+(y-1)^2=13/4`

This is the equation of a circle, center `(-3/2,1)` and radius `=sqrt13/2`

Answer 2

`2y-3x=x^2+y^2` or `y=2+-sqrt(-x^2-3x+1)`

Explanation:

You are given an equation in terms of radius, `r`, and angle, `theta`. In order to find the rectangular form, you must know the relationships between the polar coordinate system, `(r,theta)`, and the rectangular coordinate system, `(x, y)`.

• `r=sqrt(x^2+y^2)`
• `sin(theta)=y/r=y/sqrt(x^2+y^2)`
• `cos(theta)=x/r=x/sqrt(x^2+y^2)`
• `tan(theta)=y/x`

Now we can replace the `theta` and `r` in the original problem with expressions in terms of `x` and `y`.

`2sin(theta)-3cos(theta)=r`
`2y/sqrt(x^2+y^2)-3x/sqrt(x^2+y^2)=sqrt(x^2+y^2)`

Multiply both sides of the equation by `sqrt(x^2+y^2)`

`2y-3x=x^2+y^2`

Solving for `y`, you get

`y^2-2y=-x^2-3x` ...separate x's and y's
`y^2-2y+1=-x^2-3x+1` ...complete the square
`(y-2)^2=-x^2-3x+1` ...complete the square
`y-2=+-sqrt(-x^2-3x+1)` ...square root both sides
`y=2+-sqrt(-x^2-3x+1)` ...add 2 to both sides

The graph of this is a circle