Question

Tentukanlah persamaan garis singgung pada lingkaran x2+y2-4x+6y-12=0 melalui titik (6,-6)

Answer 1

2A = -4      2B = 6    C = -12
A = -2          B = 3

r =
r =
r = √25
r = 5

Pusat(-A,-B)
Pusat(2, -3)

(6,-6)
x1,y2

(x1-a)(x-a) + (y2-b)(y-b) = r^2
(6 - 2)(x-2) + (-6-(-3))(y-(-3)) = 5^2
4x - 8 - 3y - 9 = 25
4x - 3y - 42 = 0

Answer 2

Pusat ( 2, -3)
titik (6,-6)
m = -6 +3 / 6-2
m = -3 /4
m gris singgung = 4/3
pers g singgung
y-y1 = mg (x- x1)
y +6 = 4/3 (x-6 )
3y + 18 = 4 x - 24
4x - 3y - 42= 0