Question

Determine the center and radius of the following circle equation: x^2 + y^2 + 12x + 16y + 19 = 0

Answer 1

Answer:

radius: 9

center: (-6, -8)

Step-by-step explanation:

First, you may wan to rearrange the equation so it is easier to sole, like:

x^2 + 12x + y^2 + 16y + 19 =0

Next, we need to complete the square. x^2 + 12x is the start of the square of x + 6, which ends in 36, and y^2 + 16y is the start of the square of x + 8, which ends in 64. We can start to write in the squares like:

x^2 + 12x + 36 - 36 + y^2 + 16y + 64 - 64 + 19 = 0

(36 & 64 are subtracted so the equation stays the same)

You can factor getting:

(x + 6)^2 - 36 + (y + 8)^2 - 64 + 19 = 0

You can combine the constants to get -36 - 64 + 19 = -81 and we can add 81 to both sides to get:

(x + 6)^2 + (y + 8)^2 = 81

The standard form of a circle equation is:

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

where (h, k) is the center and r is the radius.

In this case, by substituting, we get the center is (-6, -8) and the radius is 9.