What is the general form of the equation of the given circle with center A? CenterA: (-3,12) Radius:5

x2 + y2 + 6x - 24y - 25 = 0

x2 + y2 - 6x + 24y + 128 = 0

x2 + y2 + 6x - 24y + 128 = 0

x2 + y2 + 6x - 24y + 148 = 0

Standard form of a circle is:

(x-h) ^2 + (y-k) ^2 = r^2

where

(h, k) is the center

and

r is the radius

.

so we have,

(x-h) ^2 + (y-k) ^2 = r^2

put the center (-3,12) and radius 5

(x - (-3)) ^2 + (y-12) ^2 = 5^2

(x+3) ^2 + (y-12) ^2 = 25 (general form)

.

expand to get "general form"

x^2 + y^2 + Dx + Ey + F = 0

(x+3) ^2 + (y-12) ^2 = 25

(x+3) (x+3) + (y-12) (y-12) = 25

x^2+6x+9 + y^2-24y+144 = 25

x^2+y^2+6x-24y+153 = 25

x^2+y^2+6x-24y+128 = 0 (answer)