一個三角形的3個頂點是A(3,2).B(-5,6) C(-3,10) 求穿過A B C 和 C 3點的圓方程

1)設圓心坐標為( h , k )。
√[ ( h – 3 )2 + ( k – 2 )2 ] = r
√[ ( h + 5 )2 + ( k – 6 )2 ] = r
√[ ( h + 3 )2 + ( k – 10 )2 ] = r
於是,
√[ ( h – 3 )2 + ( k – 2 )2 ] = √[ ( h + 5 )2 + ( k – 6 )2 ]
( h – 3 )2 + ( k – 2 )2 = ( h + 5 )2 + ( k – 6 )2
( h – 3 + h + 5 )( h – 3 – h – 5 ) = ( k – 6 + k – 2 )( k – 6 – k + 2 )
( 2h + 2 )( - 8 ) = ( 2k – 8 )( - 4 )
k = 2h + 6 --- ( 1 )
√[ ( h + 5 )2 + ( k – 6 )2 ] = √[ ( h + 3 )2 + ( k – 10 )2 ]
( h + 5 )2 + ( k – 6 )2 = ( h + 3 )2 + ( k – 10 )2
( h + 5 + h + 3 )( h + 5 – h – 3 ) = ( k – 10 + k – 6 )( k – 10 – k + 6 )
( 2h + 8 )( 2 ) = ( 2k – 16 )( - 4 )
2k + h – 12 = 0 --- ( 2 )
代( 1 )入( 2 ),
2 ( 2h + 6 ) + h – 12 = 0
5h = 0
h = 0
k = 2 ( 0 ) + 6 = 6
所以圓心坐標為( 0 , 6 )。
半徑( r ): √[ ( 0 – 3 )2 + ( 6 – 2 )2 ] = 5
圓方程: ( x – 0 )2 + ( y – 6 )2 = 25
x2 + y2 – 12y + 11 = 0
2)C1: x2 + y2 – 2x + 6y – 15 = 0
代y = 0,
x2 – 2x – 15 = 0
( x – 5 )( x + 3 ) = 0
x = 5 或 –3
所以AB 坐標分別是( 5 , 0 )和( - 3 ,0 )。
C2圓心: ( 1 , 0 ), 半徑: [ 5 – ( - 3 ) ] / 2 = 4
C2: ( x – 1 )2 + ( y – 0 )2 = 42
x2 + y2 – 2x – 15 = 0
C1圓心: ( 1 , - 3 )
代( 1 , - 3 )入C2,
L.H.S. = ( 1 )2 + ( - 3 )2 – 2 ( 1 ) – 15 = 8 – 15 = -7 ≠ 0
所以C2不穿過C1的圓心。

x^2 + y^2 - 12y + 11 = 0