Persamaan garis singgung lingkaran tegak lurus garis 2x-8y-5 = 0 menyinggung lingkaran x^2+y^2-6x+2y-6 =0 Butuh cepat ya ! Thanks minna :)

2x-8y-5 = 0
m1 = 1/4
Tegak lurus = m1 x m2 = -1 => m2 = -4

x2+y2-6x+2y-6 = 0
Titik pusat = -A/2,-B/2 = -(-6/2), -(2/2) = 3,-1

Memakai cara kuadrat sempurna :
x2+y2-6x+2y-6 = 0
x2 - 6x + (-3)^2 + y2 + 2y + (1)^2 = 6 + (-3)^2 + (1)^2
(x-3)^2 + (y+1)^2 = 6 + 9 + 1
(x-3)^2 + (y+1)^2 = 16 => r^2 = 16 => r = 4

Persamaan garis singgung dengan :
m = -4
Titik pusat (a, b) = (3,-1)
r = 4

y-b = m(x-a) ± r √1+m^2
y+1 = -4 (x-3) ± 4 √1+ (-4)^2
y+1 = -4x + 12 ± 4 √1+16
y+1 = -4x + 12 ± 4√17
y = -4x + 11 ± 4√17