Question

Diketahui persamaan lingkaran:
2X²+2Y²+kX-12Y+6=0
melalui titik (-1,4)
maka titik pusatnya adalah?

Answer 1

Verified answer

Jawab

subs (-1, 4) ke 2x^2 +2y^2 + kx -12y + 6 = 0
2(-1)^2 + 2(4)^2 + k(-1)  -12(4) +6= 0
2 + 32 - k - 48 +6 = 0
-k -8 = 0
k  = -8

2x^2  + 2y^2 -8x -12y + 6 = 0 ....bagi 2
x^2 +y^2 - 4x - 6y + 3 = 0
P(1/2 (4), 1/2 (6))
p (2, 3)