Question

数学题！！！急求

经过P（-2，4），Q（3，-1）两点，并且在X轴上截得的玄长为6。求园的方程

Answer 1

设所求圆的方程为，(x-a)^2 + (y-b)^2 = c.
则，
y = 0,
(x-a)^2 = c - b^2,
x1 = a + (c-b^2)^(1/2),
x2 = a - (c-b^2)^(1/2).
2(c-b^2)^(1/2) = 6,
c - b^2 = 9,
c = b^2 + 9.

(-2-a)^2 + (4-b)^2 = c,
(3-a)^2 + (-1-b)^2 = c.

4 + 4a + a^2 + 16 - 8b + b^2 = 9 - 6a + a^2 + 1 + 2b + b^2,
10a - 10b + 10 = 0,
a = b - 1.

[3-(b-1)]^2 + (-1-b)^2 = b^2 + 9,
(4-b)^2 + (1+b)^2 = b^2 + 9,
16 - 8b + b^2 + 1 + 2b + b^2 = b^2 + 9,
b^2 - 6b + 8 = 0,
(b-2)(b-4) = 0,
b = 2,或者，b = 4.
a = 1,或者，a = 3.
c = 13,或者，c = 25.

所求圆的方程为，(x-1)^2 + (y-2)^2 = 13,或者，(x-3)^2 + (y-4)^2 = 25.