Question

Persamaan garis singgung lingkaran x²+y²-2x-4y-4=0 di titik (0,5) adalah

Answer 1

Verified answer

Jawab

x²+y² -2x - 4y  - 4= 0
(x -1)² +(y-2)² = 9 ...(i)

titik (0,5) di luar lingkaran
garis kutub di (0,5)
(0-1)(x -1)+ (5-2)(y-2) = 9
-x + 1 + 3y - 6 = 9
-x + 3y -5 = 9
-x + 3y = 14
x = 3y -14 ...(ii) sub ke lingkaran (i)
(3y-14 -1)² +(y-2)² =9
(3y-15)² + (y-2)² = 9
9y² - 90y + 225 +y² - 4y + 4 = 9
10y² - 94 y + 220 = 0
5y² - 47 y + 110 = 0
(5y -  22)(y - 5)= 0
y = 22/5  atau y = 5
x = 3y - 14
y = 22/5 --> x = -4/5
y = 5 --> x = 1
titiksinggung (1, 5)  atau ( -4/5, 22/5)

PGS  (y -y1)/(y2-y1)  = ( x- x1)/(x2 - x1)

1) dititik (0,5) dan (1,5)
(y-5)/(0) = (x - 0)/(1-0)
y - 5 = 0
y = 5

Answer 2

Verified answer

Jawab :

Beda cara ..

Cek titik thd lingkaran :
(0,5)
0² + 5² - 2.0 - 4.5 - 4
= 25 - 24
= 1
titik diluar lingkaran

Persamaan garis dg gradien m dan mlalui (0,5) :
y = mx + 5

x² + y² - 2x - 4y - 4 = 0
x² + (mx + 5)² - 2x - 4(mx + 5) - 4 = 0
(1 + m²)x² + (6m - 2)x + 1 = 0

Garis menyinggung lingkaran : D = 0
D = 0
(6m - 2)² - 4(1 + m²) = 0
32m² - 24m = 0
4m² - 3m = 0
m(4m - 3) = 0
m = 0 atau m = 3/4

Utk m = 0
PGSL :
y = mx + 5
y = 5 ✓

Utk m = 3/4
PGSL :
y = 3/4 x + 5
3x - 4y + 20 = 0 ✓