Tentukan persamaan garis singgung lingkaran berikut. a) L= x^2 + y^2-4x+6y-7=0 tegak lurus garis x+2y=7 b) (x-3)^2 + (y+2)^2=25 bergradien √3

a) L = x²+y²-4x+6y-7 = 0, tegak lurus x+2y=7
A=-4 B=6 C=-7
O(a,b) = (-A/2, -B/2) = (2,-3)
r = √a²+b²-C = √20
m = -1/2, karena tegak lurus maka,
mg×m = -1 → mg = -1/-½ = 2.

y-b = mg(x-a) ± r√mg²+1
y+3 = 2(x-2) ± √20.√5
y+3 = 2x-4 ± 10
Pers. garis singgungnya :
• 2x-y+3 = 0
• 2x-y-17 = 0

b) (x+3)²+(y+2)² = 25, gradien √3
x²+6x+9+y²+4y+4 = 25
x²+y²+6x+4y-12 = 0
A=6 B=4 C=-12
O(a,b) = (-A/2, -B/2) = (-3,-2)
r = √a²+b²-C = √-3²+-2²+12 = 5
m = √3

y-b = m(x-a)±r√m²+1
y+2 = √3(x+3)±5√(√3)²+1
y+2 = √3.x +3√3 ± 10
pers. garis singgungnya :
• x√3 - y + 8+3√3 = 0
• √3 - y - 12+3√3 = 0