pusat dan jari-jari lingkaran x^2 + y^2 - 10x + 14y + 58 = 0 adalah

x2 + y2 -10x + 14y + 58 = 0

Titik pusat : (a, b) => (5,-7)
a = -A/2 = -(-10/2) = 5
b = -B/2 = -(14/2) = -7

jari jari : Mencari dg persamaan kuadrat
x2 - 10x + (-5)^2 + y2 + 14y + (7)^2 = -58 + (-5)^2 + (7)^2
(x-5)^2 + (y+7)^2 = -58 + 25 + 49
(x-5)^2 + (y+7)^2 = 16
r^2 = 16 => r = 4