1.The diameters of a circle are along 2x+y−7=02x+y-7=02x+y−7=0 and x+3y−11=0x+3y-11=0x+3y−11=0. Then, the equation of this circle, which also passes through (5,7)(5,7)(5,7), isa.x2+y2−4x−6y−16=0x^2+y^2-4x-6y-16=0x2+y2−4x−6y−16=0b.x2+y2−4x−6y−20=0x^2+y^2-4x-6y-20=0x2+y2−4x−6y−20=0c.x2+y2−4x−6y−12=0x^2+y^2-4x-6y-12=0x2+y2−4x−6y−12=0d.x2+y2+4x+6y−12=0x^2+y^2+4x+6y-12=0x2+y2+4x+6y−12=0Login to continueOnly logged in users canattempt or see the solution.
