Question

How do you convert `r=12 cos [theta]` to rectangular form?

Answer 1

The equation becomes a circle with radius 6 and centered at `(6, 0)`: `(x - 6)^2 + (y - 0)^2 = 6^2`

Explanation:

Multiply both sides by r:

`r^2 = 12rcos(theta)`

Substitute `x^2 + y^2` for `r^2` and x for `rcos(theta)`

`x^2 + y^2 = 12x`

Subtract 12x from both sides:

`x^2 - 12x + y^2 = 0`

Add h^2 to both sides:

`x^2 - 12x + h^2 + y^2 = h^2`

Complete the square for #(x - h)^2 = x^2 - 2hx + h^2:

-2hx = -12x

`h = 6`

Insert the square term with h = 6 on the left and substitute 6 for h on the right:

`(x - 6)^2 + y^2 = 6^2`

Write the y term is standard form:

`(x - 6)^2 + (y - 0)^2 = 6^2`

Answer 2

`x^2+y^2-12x= 0`

Explanation:

Multiply the given expression on both sides by r, it becomes `r^2= 12 r cos theta`

Substitute `rcos theta =x` and `r^2= x^2+y^2`

The equation now becomes, `x^2+y^2 = 12x`
`x^2 +y^2 -12x=0`