F5 math problems ( Equation of Circles )

Let (h,k) be the centre of circle
The centre of a circle lies on the straight line 2x + y - 4 = 0
2h + k - 4 = 0------------------------(1)

the circle passes through A( 0, 1) and B( 2, -1)
h² + (k-1)² = (h-2)² + (k+1)² = square of radius
h² - (h-2)² = (k+1)² - (k-1)²
(2h-2)(2) = (2k)(2) [By a²-b²≡a²+2ab+b²]
h - 1 = k--------------------------(2)

sub (2) into (1):
2h + (h-1) - 4 = 0
3h = 5
h = 5/3

sub h = 5/3 into (2):
k = 5/3 - 1 = 2/3

∴centre =

equation of the circle:
(x - 5/3)² + (y- 2/3)² = h² + (k-1)²
(x - 5/3)² + (y- 2/3)² = (5/3)² + (2/3 - 1)²
(x - 5/3)² + (y- 2/3)² = 26/9
(3x - 5)² + (3y- 2)² = 26

2007-11-08 15:57:35 補充：
centre = (5/3 , 2/3)

2007-11-08 16:52:32 補充：
equation of the circle 是 (x - 5/3)² (y- 2/3)² = h² (k-1)²因為equation of any real circle是(x - h)² (y - k)² = r²(h,k)是圓心,r是半徑

Let the equation of the circle is:
(x-A)2 + (y-B)2 = a^2 …… ( The standard form of circle and (A, B) is the center of the circle, a is the radius)
Put (0, 1) into the equation
(0-A)2 + (1-B)2 = a^2
=> A2 + 1 – 2B + B2 = a^2 …………………..(i)
Put (2, -1) into the equation
4 – 4A + A2 + 1 + 2B + B2 = a^2…………….(ii)
(ii) – (i)
1 + B = A
Because the center of circle (A, B) lies on the equation 2x + y – 4 = 0
so put (1 + B, B) into the equation 2x + y -4 = 0
2 + 2B + B – 4 = 0
3B = 2
B = 2/3
so A = 5/3
put the result of A and B into the (i)
a^2 = 26/9
than the equation of the circle is
3x^2 + 3y^2 – 10x – 4y + 1 = 0

2007-11-07 19:12:41 補充：
The unknown A,B are not the point A and B in the question.