Question

Persamaan garis singgung lingkaran x^2+y^2-26x+8y+160=0 yang tegak lurus garis 12x+5y=10 adalah?

Answer 1

12x + 5y = 10
gradien = m1 = -12/5
tegak lurus m1.m2 = -1
-12/5 . m2 = -1
m2 = 5/12

x² + y² - 26x + 8y + 160 = 0
(x - 13)² + (y + 4)² - 25 = 0
(x - 13)² + (y + 4)² = 25
(x - 13)² + (y+4)² = 5²

y + 4 = 5/12(x - 13) ± 5√(1+(5/12)²)
y + 4 = 5/12(x - 13) ± 5√(1+(25/144))
y + 4 = 5/12(x - 13) ± 5√(169/144)
y + 4 = 5/12(x - 13) ± 5(13/12)
kalikan 12
12y + 48 = 5(x - 13) ± 5(13)
12y + 48 = 5x - 65 ± 65
12y - 5x = 48 - 65 ± 65
12y - 5x = 48 - 65 + 65
12y - 5x = 48
atau
12y - 5x = 48 - 65 - 65
12y - 5x = -82