persamaan garis singgung pada lingkaran x² + y² - 2x + 4y - 4 = 0 yang tegak lurus garis 5x - 12y + 15 = 0

x² + y² - 2x + 4y - 4 = 0
A = -2, B = 4 dan C = -4
pusat = (-A/2, -B/2) = (1, -2)
r = √(1² + (-2)² - C) = √(1 + 4 + 4) = 3
5x - 12y + 15 = 0 → 12y = 5x + 15
y = (5/12)x + 15/12 → m = 5/12
tgk lurus m1 = -1/m = -1/(5/12) = -12/5
y - b = m1(x - a) ± r√(1 + m1²)
y + 2 = -(12/5)(x - 1) ± 3√(1 + (-12/5)²)
y + 2 = -(12/5)(x - 1) ± 39/5, dikali 5
5y + 10 = -12(x - 1) ± 39
PGS 1 dan 2
5y = -12x + 2 ± 39