tentukan persamaan garis yg melalui titik (12,5) dan menyinggung lingkaran x²+y²-14x-10y+58=0

a = -1/2 (-14) = 7
b = -1/2 (-10) = 5
r2 = (7-12)^2 + (5-5)^2
= (-5)^2 = 25
(x1-a)(x-a) + (y1-b)(y-b) = r^2
(12-7)(x-7) + (5-5)(y-5) = 25
5 (x-7) = 25
5x-35 = 25
------------------:5
x-7 = 5
x-12 = 0