Find the equation of the diameter of the circle x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0x2+y2−6x+4y−12=0 that passes through (5,2)(5, 2)(5,2).
2x−y−8=02x - y - 8 = 02x−y−8=0
x−2y+1=0x - 2y + 1 = 0x−2y+1=0
x+2y−9=0x + 2y - 9 = 0x+2y−9=0
2x+y−12=02x + y - 12 = 02x+y−12=0