What is the equation of a circle with center (-8, 3) and radius 8?

x^2 + y^2 + 16x + 6y + 9 = 0

Step-by-step explanation:

Using the formula for equation of a circle

(x - a) ^2 + (y + b) ^2 = r^2

(a, b) - the center

r - radius of the circle

Inserting the values given in the question

(-8,3) and r = 8

a - - 8

b - 3

r - 8

[ x - (-8) ]^2 + (y+3) ^2 = 8^2

(x + 8) ^2 + (y + 3) ^2 = 8^2

Solving the brackets

(x + 8) (x + 8) + (y + 3) (y+3) = 64

x^2 + 16x + 64 + y^2 + 6y + 9 = 64

Rearranging algebrally,.

x^2 + y^2 + 16x + 6y + 9+64 - 64 = 0

Bringing in 64, thereby changing the + sign to -

Therefore, the equation of the circle =

x^2 + y^2 + 16x + 6y + 9 = 0