Persamaan garis singgung lingkaran x²+y²-4x+6y-12=0 di titik (2,2) adalah
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Cara kerja
x² + y² - 4x + 6y - 12 = 0
A = -4
B = 6
C = -12
x1 = 2
y1 = 2
Persamaan garis singgung lingkaran
xx1 + yy1 + A/2 (x + x1) + B/2 (y + y1) + C = 0
2x + 2y + (-4)/2 (x + 2) + 6/2 (y + 2) - 12 = 0
2x + 2y - 2x - 4 + 3y + 6 - 12 = 0
5y - 10 = 0
5y = 10
y = 2
Mapel : Matematika
Bab : Lingkaran
Kode : 11.2.5.1