Answer:
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
