Persamaan garis singgung dari titi(0,0) pada lingkaran (x-3) pangkat dua + (y -4) pangkat dua =5 adalah...

Garis singgung pada lingkaran

(x-3)^2 + (y-4)^2 = 5 dititik (0,.0)
(0-3)(x-3) + (0-4)(y-4) = 5
-3x+9  - 4y +16 = 5
-3x-4y = -20
3x + 4y = 20
4y = 20 - 3x
y = 5 - 3/4 x

(x -3)^2 + (5 - 3/4 x - 4)^2 = 5
(x-3)^2  +(1 - 3/4 x)^2 = 5
x^2 -6x + 9 + 1 - 3/2x + 9/16 x^2 =  5
x^2 + 9/16x^2  - 6x - 3/2 x + 5 = 0

16x^2 + 9x^2 - 96x - 24x + 80 = 0
25x^2 - 120x + 80  = 0
5x^2 - 24x + 16 = 0
(5x - 4)(x -  4)  = 0
x= 4/5, y = 4 2/5
x = 4 , y = 2

garis singgung
di (4,2) --> ((4-3)(x-3) + (2-4)(y-4) = 5
x-3 -2y+8-5=0
x - 2y = 0
2y = x
y = 1/2 x
.
di(4/5 , 4 2/5) --> (4/5 - 3)(x-3) + (4 2/5 -4)(y-4)= 5
-11/5(x-3) + 2/5 (y-4)= 5
-11(x-3) + 2(y-4)= 25
-11x + 33 + 2y -8 - 25 =0
-11x + 2y = 0
2y = 11x
y =  11/2  x