Equation

2y = b … (1)
2x = a + m … (2)
a2 + b2 = 4 … (3)
(m – a)2 + b2 = 81 … (4)
(2) => m = 2x – a
Sub into (4), (2x – a – a)2 + b2 = 81
4x2 – 8ax + 4a2 + b2 = 81
Use (3), 4x2 – 8ax + 4(4 – b2) + b2 = 81
4x2 – 8ax + 16 – 3b2 = 81
a = [4x2 – 3(2y)2 – 65] / (8x) since b = 2y
a = (4x2 – 12y2 – 65) / (8x)
Sub into (3),
(4x2 – 12y2 – 65)2 / (8x)2 + (2y)2 = 4
(4x2 – 12y2 – 65)2 = (8x)2(4 – 4y2)
16x4 + 144y4 + 4225 – 96x2y2 – 520x2 + 1560y2 = 256x2 – 256x2y2
16x4 + 144y4 + 160x2y2 – 776x2 + 1560y2 + 4225 = 0
By solving the above quadratic equation in y2 alternative form is
y2 = [+/-8x√(4x2 + 231) – (20x2 + 195)] / 36.
Graph below:

圖片參考：http://img511.imageshack.us/img511/2806/locus.jpg

I agree 002 is wrong. But I am also doubtful about your answer, I know 5 equations are required to solve 5 unknowns, but why we cannot find a relation , not answer, to x and y with 4 equations.

002回答是錯的。
有4條equations但5個unknowns，是有無限多組解，並不只局限於一組解。
而且，題目沒說明variables是整數，所以他錯。

If a, b, m, x, y都係整數

(4)-(3):
(m-a)^2 - a^2 = 77
m^2 -2am + a^2 - a^2 =77
m(m-2a)=77

Since 77=11*7
So m =11, a=2 : 11*(11-4)=77

Put a , m 代入(2)
2x=a+m
2x=18
x=9

Put a 代入(3)

a^2 + b^2 =4
4+b^2=4
b=0

Sinceb=0,y=0

So:a=2, b= 0, m=11, x=9 , y =0

You can say x=y+9