Question

Persamaan garis singgung lingkaran x²+y²-6x+4y-12=0 di titik(7,1) adalah..

Answer 1

(x₁, y₁) = (7, 1)

x² + y² - 6x + 4y - 12 = 0  ⇒ A = -6  B = 4  C = -12
x₁x + y₁y + A  + B + C = 0
(7)x + (1)y - 6 + 4 - 12 = 0
14x + y - 42 - 6x + 4 + 4y - 12 = 0
8x + 5y - 52 = 0