Question

Jika garis y=2x-4 memotong lingkaran x^2+y^2+6x-20=0 di titik K dan L, panjang tali busur KL adalah
A.2√6
B.4
C.6
D.6√2
E.6√3

Answer 1

Verified answer

Y = 2x - 4
x^2 + y^2 + 6x - 20 = 0
x^2 + (2x - 4)^2 + 6x - 20 = 0
x^2 + 4x^2 - 16x + 16 + 6x - 20 = 0
5x^2 - 10x - 4 = 0
x = (-b ± √(b^2 - 4ac))/(2a)
= (10 ± √(100 + 80)) / 10
= (10 ± √180) / 10
= (10 ± 6√5) / 10
x1 = (5 + 3√5)/5 atau x2 = (5 - 3√5)/5

y = 2x - 4
y1 = 2(5 + 3√5)/5 - 4 = (10 + 6√5 - 20)/5 = (6√5 - 10)/5
atau
y2 = 2(5 - 3√5)/5 - 4 = (10 - 6√5 - 20)/5 = (-6√5 - 10)/5

(x2 - x1) = (5 - 3√5)/5 - (5 + 3√5)/5 = -6√5 / 5
(y2 - y1) = (-6√5 - 10)/5 - (6√5 - 10)/5 = -12√5 / 5

Panjang busur KL
= √((x2 - x1)^2 + (y2 - y1)^2)
= √((-6√5/5)^2 + (-12√5/5)^2)
= √(36 . 5 / 25 + 144 . 5 / 25)
= √(36/5 + 144/5)
= √(180/5)
= √36
= 6