Question

What is the center of the circle X2 + y2 = 28+ 3x?

Answer 1

The center is
`((3)/(2),0)`.

The problem:

What is the center of the circle X2 + y2 = 28+ 3x?

Explanation:

We need to put anything with an
`x` together and anything with a
`y` together:

`x^2+y^2=28+3x`

Subtract
`3x` on both sides:

`(x^2+y^2)-3x=(28+3x)-3x`

`x^2-3x+y^2=28`

We need to complete the square for
`x`'s. The
`y`'s are already done.

Keep in mind:
`x^2+bx+((b)/(2))^2=(x+(b)/(2))^2</p><p></p><p>So we are going to add [tex]((-3)/(2))^2` on both sides so we can use this identity for the left side with the
`x`'s:

`x^2-3x+((-3)/(2))^2+y^2=28+((3)/(2))^2`

`(x+(-3)/(2))^2+y^2=28+(9)/(4)`

`(x-(3)/(2))^2+y^2=(28(4)+9)/(4)`

`(x-(3)/(2))^2+(y-0)^2=(121)/(4)`

So the center is
`((3)/(2),0)`.

The radius is
`\sqrt{(121)/(4)}=(11)/(2)`.