The idea here is to consider the line containing the origin and the center of the circle, and find the intersection of this line with the given circle. You will get two points, one which closest to the origin and the other which is farthest from the origin.

Rewrite the equation of the circle in standard form by completing the square.

x^{2 }- 12x + 36 + y^{2 }- 4y + 4 = 50 + 36 + 4

(x - 6)^{2 }+ (y - 2)^{2 }= 90

The center of the circle (6,2), so the line containing (6,2) and (0,0) is y = (1/3)x. The intersection of the line and circle are found by solving the equations simultaneously. Substitute (1/3)x for y in the equation x^{2} + y^{2} - 12x - 4y = 50 to get

x^{2 }+ (x/3)^{2 }- 12x - 4x/3 = 50

x^{2 }+ x^{2}/9 - 12x - 4x/3 = 50

10x^{2}/9 - 40x/3 - 50 = 0

x^{2 }- 12x - 45 = 0 [multiply the previous equation by 9/10]

(x - 15)(x + 3) = 0

x = 15, -3

y = 5, -1

The points of intersection are (15, 5) and (-3, -1). The point closest to the origin is (-3, -1) and (15, 5) is the point farthest from the origin.