Tentukan persamaan garis singgung pada lingkaran x²+y²-2x+4y-4=0 yang tegak lurus garis 5x-12y+5=0! (Pake cara yaa)? Terima kasih

x² + y² - 2x + 4y - 4 = 0 => A = -2, B = 4, C = -4
Pusat = (a, b) = (-1/2 A, -1/2 B) = (-1/2 (-2), -1/2 (4)) = (1, -2)
Jari-jari = r = √(a² + b² - C) = √(1² + (-2)² - (-4)) = √(1 + 4 + 4) = √9 = 3

5x - 12y + 5 = 0 => m = -x/y = -5/-12 = 5/12 karena tegak lurus m = -12/5
(Ingat tegak lurus berlaku rumus m1.m2 = -1)

Persamaan garis singgung lingkaran
(y - b) = m (x - a) ± r √(m² + 1)
y - (-2) = -12/5 (x - 1) ± 3 √((-12/5)² + 1)
y + 2 = -12/5 x + 12/5 ± 3 √(144/25 + 25/25)
y + 2 = -12/5 x + 12/5 ± 3 √(169/25)
y + 2 = -12/5 x + 12/5 ± 3 . 13/5 ============> kedua ruas kali 5
5y + 10 = -12x + 12 ± 39
12x + 5y + 10 - 12 ± 39 = 0
1) 12x + 5y - 2 + 39 = 0 => 12x + 5y + 37 = 0
2) 12x + 5y - 2 - 39 = 0 => 12x + 5y - 41 = 0

Lingkaran.
Kelompok peminatan kelas XI kurikulum 2013 revisi 2016.

x² + y² + Ax + By + C = 0
x² + y² - 2x + 4y - 4 = 0

5x - 12y + 5 = 0
m₁ = -5 / -12 = 5 / 12
m₁ m₂ = -1
m₂ = -1 / m₁ = -1 / (5 / 12) = -12 / 5

r = √(1/4 A² + 1/4 B² - C)
   = √[1/4 (-2)² + 1/4 (4)² + 4] = 3

y + 1/2 B = m₂(x + 1/2 A) ± r √(1 + m₂²)
y + 2 = -12/5 (x - 1) ± 3 √[1 + (-12/5)²]
y = -12/5 (x - 1) ± 39/5 - 2
5y = -12 (x - 1) ± 39 - 10
5y = -12x + 41 dan 5y = -12x - 37
12x + 5y - 41 = 0 dan 12x + 5y + 37 = 0