Question

Find an equation of the circle passing
through (9,-7)(-3,1)&(6,2)​

Answer 1

Answer:

Let the circle be

${x}^{2} + {y}^{2} + dx + ey + f = 0$

(9,7)

$81 + 49 + 9d + 7e + f = 0 \\ 9d + 7e + f = - 130$

(-3,1)

$9 + 1 - 3d + e + f = 0 \\ - 3d + e + f = - 10$

(6,2)

$36 + 4 + 6d + 2e + f = 0 \\ 6d + 2e + f = - 40$

Solving the equation

We get,

$d = - \frac{10}{7} \\ e = - \frac{120}{7} \\ f = \frac{20}{7}$

So, the equation of circle is,

${x}^{2} + {y}^{2} - \frac{10}{7}x - \frac{120}{7}y + \frac{20}{7} = 0$