Question

A circle has an equation 4(x-2)^2+4y^2. Find the center and radius. This is not in general form and I am not sure hot to convert to standard form. Can you assist step by step?

Answer 1

See a solution process below:

Explanation:

The equation for a circle is:

`(x - color(red)(a))^2 + (y - color(red)(b))^2 = color(blue)(r)^2`

Where `(color(red)(a), color(red)(b))` is the center of the circle and `color(blue)(r)` is the radius of the circle.

We can convert the formula in the problem to the general formula for a circle as follows:

`4(x-2)^2 + 4y^2=4`

Divide each side of the equation by `color(red)(4)` to put the equation in the general form of an equation for a circle:

`(4(x-2)^2 + 4y^2)/color(red)(4) = 4/color(red)(4)`

`(4(x-2)^2)/color(red)(4) + (4y^2)/color(red)(4) = 1`

`(color(red)(cancel(color(black)(4)))(x-2)^2)/cancel(color(red)(4)) + (color(red)(cancel(color(black)(4)))y^2)/cancel(color(red)(4)) = 1`

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

`(x - color(red)(2))^2 + (y - color(red)(0))^2 = color(blue)(1)^2`

Center of the Circle is: `(2, 0)`

Radius of the Circle is:

`r^2 = 1`

`sqrt(r^2) = sqrt(1)`

`r = 1`