Persamaan garis singgung lingkaran x^2 + y^2 - 6x + 4y = 12 yang tegak lurus garis 4x - 3y = 5 adalah

A = -6, B = 4 da C = 0
pusat = (-A/2, -B/2) = (3, -2)
r = √(3² + (-2)² - C) = √(9 + 4 - 0) = √13
3y = 4x - 5 → y = (4/3)x - 5/3 → m = 4/3
tgk lurus → m1 = -1/m = -1/(4/3) = -3/4
y - b = m1(x - a) ± r√(1 + m1²)
y + 2 = -(3/4)(x - 3) ± √13√(1 + (-3/4)²)
y = -(3/4)x + (1/4) ± (5/4)√13, atau
4y = -3x + 4 ± 5√13

pusat lingkaran (-A/2, -B/2)
(6/2, -4/2)
(3,-2)
M sejajar =-a/b
=-4/-3
=4/3 M tegak lurus = -3/3
= Y -B = M(x-a)
=plus minus r akar 1 + M^2
=Y+2=-3/4(x-) plus minus akar 2. akar 1 +(-3/4)^2
=y+2=-3/4x+9/4 plus minus 5/4 akar 12
=y=-3/4x +1/4x +1/4+5/4akar12
=y=-3/4x+1/4-5/4 akar 12