Question

find the equation of the circle having (7,9) and (11,-7) as end of it's diameter​

Answer 1

Answer:

${x}^{2} + {y}^{2} - 18x - 16y + 140 = 0.$

Step-by-step explanation:

$(x - 7)(x - 11) + (y - 9)(y + 7) = 0 \\ \\ {x}^{2} + {y}^{2} - 18x - 16y + 77 + 63 = 0 \\ \\ {x}^{2} + {y}^{2} - 18x - 16y + 140 = 0.$