數學問題1條(須列式)(15分)

Let the x- coordinate of the centre of the circle be a.
So the y - coordinate of the centre of the circle will be -a since the centre is on line x+y = 0. Therefore,
(a - 1)^2 + (-a -2)^2 = (a - 5)^2 + (-a - 0)^2.
a^2 + 1 - 2a + a^2 + 4 + 4a = a^2 + 25 - 10a + a^2
5 + 2a = 25 - 10a
12a = 20
a = 5/3. So the centre is (5/3, -5/3).
Radius =sqrt[ (5/3 - 5)^2 + (5/3)^2 ]= sqrt(100/9 + 25/9) = (sqrt125)/3.
So equation of circle is
(x - 5/3)^2 + (y +5/3)^2 = 125/9.
x^2 + 25/9 - 10x/3 + y^2 + 25/9 + 10y/3 = 125/9
x^2 + y^2 - 10x/3 + 10y/3 - 25/3 = 0

圓心位於直線 x+y=0 上
故設圓心O = (h,-h)
AO = OB
(1-h)^2 + (2+h)^2 = (5-h)^2 + (0+h)^2
1-2h+h^2+4+4h+h^2 = 25-10h+h^2+h^2
5+2h+2h^2 = 25-10h+2h^2
12h = 20
h = 5/3
O = (5/3,-5/3)
半徑 = sqrt[(1-5/3)^2+(2+5/3)^2] = sqrt(125/9) = 5(sqrt5)/3
方程 : (x-5/3)^2 + (y+5/3)^2 = 125/9
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