persamaan garis singgung x2+y2= 36 sejajar garis 3x+4y+20=0?

Pe_Ge_eS_eL

Lingkaran
x² + y² = 36
P(0,0)
r = √36 = 6

Garis
3x + 4y + 20 = 0
m1 = -a/b = -3/4

sejajar : m1 = m2 = m = -3/4

PGSL dg P(0,0) , m = -3/4 dan r = 6 :
y = mx ± r√(1 + m²)
y = -3/4 x ± 6√(1 + 9/16)
y = -3/4 x ± 6 . 5/4
y = -3/4 x ± 15/2

PGSL_1
y = -3/4 x + 15/2
y = 1/4 (-3x + 30)
4y = -3x + 30
3x + 4y - 30 = 0

PGSL_2
y = -3/4 x - 15/2
y = 1/4 (-3x - 30)
3x + 4y + 30 = 0