Question

1.Tentukan persamaan garis singgung pada
lingkaran L ≡ x^2 + y^2 + 6x -2y + 6 = 0
yang tegak lurus garis 3y - 4x -7 = 0

Answer 1

X^2 + y^2 + 6x - 2y + 6 = 0
pusat = (a,b) = (A/-2, B/-2) = (6/-2, -2/-2) = (-3, 1)
jari-jari = r = √(a^2 + b^2 - C) = √((-3)^2 + 1^2 - 6) = √4 = 2

3y - 4x - 7 = 0 => m = -a/b = -(-4)/3 = 4/3 karena tegak lurus m = -3/4

pers garis singgung
y - b = m (x - a) ± r √(m^2 + 1)
y - 1 = -3/4 (x - (-3)) ± 2 √((-3/4)^2 + 1)
y - 1 = -3/4 (x + 3) ± 2 √(9/16 + 1)
y - 1 = -3/4 x - 9/4 ± 2√(25/16)
y - 1 = -3/4 x - 9/4 ± 2 . 5/4 ...................... kali 4
4y - 4 = -3x - 9 ± 10
3x + 4y - 4 + 9 ± 10 = 0
1) 3x + 4y - 4 + 9 + 10 = 0
=> 3x + 4y + 15 = 0
2) 3x + 4y - 4 + 9 - 10 = 0
=> 3x + 4y - 5 = 0