Tentukan persamaan garis singgung lingkaran berikut

Klo absis = x

Klo oordinat = y

Berarti

a. x = -2

(-2)² + y² = 16

y² + 4 = 16

y² = 12

y = √12

y = 2√3

Pers garis singgung =

x.x1 + y.y1 = r²

x.-2 + y.2√3 = 16

-2x + 2√3y = 16

2x -2√3y = -16

yg b

Ordinat = -4

(x-5)² + (-4+2)² = 5

x² -10x + 25 + 4 = 5

x² – 10x + 29 = 5

x² -10x + 24 = 0

(x-4) (x-6) = 0

x = 4 atau x = 6

Jadi titiknya ada 2

(4,-4) dan (6,-4)

Klo (4,-4), pers garis singgungnya =

(x – 5)² + (y + 2)² – 5 = 0

(x-5) (a-5) + (y+2) ( b+2) -5 = 0

(x – 5)(4- 5) + (y + 2)(-4 + 2) – 5= 0

(x – 5)(-1) + (y+2)(-2) – 5 = 0

-x + 5 -2y -4 – 5 = 0

-x – 2y – 4 = 0

x + 2y + 4 = 0

Klo titik (6,-4) =

Lanjutin sendiri