Question

How do you convert `(x-1)^2 +y^2 =5` in polar form?

Answer 1

r = cos `theta` + `sqrt` ( 4 + cos `theta`. cos `theta`)

Explanation:

x = r cos `theta` and y = r sin `theta` transforms the equation to a quadratic in r. r is set as non-negative, with [0, 2`pi`] for .`theta`.
This is the polar equation of the circle with center in polar (1. 0 )and radius `sqrt`5. .

Answer 2

`r=cos theta+-sqrt(cos^2 theta+4)`

Explanation:

From the given, `(x-1)^2+y^2=5`
Use `x=r cos theta` and `y=r sin theta`

Substitute these in the given variables x and y

`(x-1)^2+y^2=5`

`(r cos theta-1)^2+(r sin theta)^2=5`

`r^2 cos^2 theta-2 r cos theta+1+r^2 sin^2 theta=5`

factor out `r^2`

`r^2(cos^2 theta+sin^2 theta)-2 r cos theta+1-5=0`
simplify
`r^2(1)-2 r cos theta-4=0`
`r^2-2 r cos theta-4=0`

Use now Quadratic Equation to solve for `r` in terms of the other variable

`r=(-b+-sqrt(b^2-4*a*c))/(2*a)`

using `a=1`, and `b=-2 cos theta`,and `c=-4`

`r=(-(-2 cos theta)+-sqrt((-2 cos theta)^2-4*1*(-4)))/(2*1)`

`r=(2cos theta+-sqrt((4 cos^2 theta+16)))/2`

which simplifies to

`r=cos theta+-sqrt(cos^2 theta+4)`

have a nice day!