Question

Select the correct answer. A circle with radius 5 and center A. The coordinates of the center of the circle are (-3, 12). What is the general form of the equation of the given circle with center A? A. x2 + y2 + 6x − 24y − 25 = 0 B. x2 + y2 − 6x + 24y + 128 = 0 C. x2 + y2 + 6x – 24y + 128 = 0 D. x2 + y2 + 6x − 24y + 148 = 0

Answer 1

you'll want to work with the center-radius form of a circle equation for this. the center formula is
`(x - h)^2 + (y - k)^2 = r^2`, where
`(h, k)` is your center and r is your radius. plug in the information your circle gives you:
`A(-3, 12)`,
`\text{radius} = 5`

`(x + 3)^2 + (y - 12)^2 = (5)^2` ... simplify the right side

`(x + 3)^2 + (y - 12)^2 = 25` ... from here, you need to foil both of your binomials to convert this to the "general form" that your answer choices are in.

`(x + 3)^2 = (x + 3)(x + 3) = x^2 + 6x + 9`

`(y - 12)^2 = (y - 12)(y - 12) = y^2 - 24y + 144`

`x^2 + 6x + 9 + y^2 - 24y + 144 = 25` ... combine like terms

`x^2 + 6x + y^2 - 24y + 153 = 25` ... subtract 25

`x^2 + 6x + y^2- 24y + 128 = 0` is your equation. reorder it so that it's from the highest degree to the lowest:

`\boxed{\bold{x^2 + y^2 + 6x - 24y + 128 = 0}}`