F5 math problems ( Equation of Circles )

圖片參考：http://quickimg.com/uploads/55861debf10b11c6fd2aac893c0e45c9.jpg

The mid - pt of ( 4,4 ) and ( 8 , 4 ): ( 6 , 4 )
Then the centre of the circle: [ 6 , ( 4 + 2 ) ] = ( 6 , 6 )
The radius of the circle = 2cm
Hence the equation: ( x - 6 )^2 + ( y - 6 )^2 = 2^2
x^2 - 12x + 36 + y^2 - 12y + 36 = 4
x^2 + y^2 - 12x - 12y + 68 = 0

2007-11-17 00:12:37 補充：
Some additions: The mid - pt of ( 4, 4 ) and ( 8 , 4 ) is found as follows:[ ( 4 + 8 ) / 2 , ( 4 + 4 ) / 2 ] = ( 6 , 4 ) It is given that the radius is 2cm, so the coordinates of the centre is [ 6, ( 4 + 2 )] = ( 6 , 6 ).

The circle is at the first quadrant.
The radius of the circle is 2. The centre is higher than the base 2 cm
The y coordinate of the centre is 6.
The x coordinate of the centre = (8+4)/2 = 6
Therefore, the coordinates of the centre = (6,6)
The equation of the circle:
(x - 6)^2 + (y - 6)^2 = 2^2
x^2 + y^2 - 12x - 12y + 68 = 0