A . Maths

a) Equation of common chord is
x^2 + y^2 -2x +3y -6 -x^2 -y^2 +4x -4y +7 = 0. That is
2x -y +1 = 0.
b) Equation of the family of circles is:
x^2 + y^2 -2x +3y -6 +k(x^2 + y^2 -4x +4y -7) = 0. That is
(1+k)x^2 + (1+k)y^2 -(2 +4k)x +(3 +4k)y -(6 +7k) = 0...................(1)
Centre is [(1+2k)/(1+k), -(3+4k)/2(1+k)]................................(2)
Square of radius is [(1+2k)/(1+k)]^2 + [(3+4k)/2(1+k)]^2 + (6+7k).....(3)
bi) Substitute (1,-2) into (1), we get
(1+k) +4(1+k) -2 -4k -6 -8k -6 -7k = 0
-14k -7 = 0, so k=-1/2. therefore equation of circle is
x^2 + y^2 +2y -5 = 0.
bii)Since centre is on x+y=7, from (2) above, that is
(1+2k)/(1+k) -(3+4k)/2(1+k) = 7
2 + 4k - 3 - 4k = 14 +14k
k=-15/14. Substitute into (1), equation of circle is
-x^2 -y^2 +16x/7 -9y/7 +3/2 = 0
14x^2 + 14y^2 -32x +18y -21 = 0.
biii) Since area = 6pi, square of radius = 6, that is equation (3) = 6. Solve for k to get equation of circles.

a) 將 C1 和 C2 兩式相等 ,
得出 x^2 + y^2 -2x +3y - 6 - x^2 - y^2 + 4x - 4y + 7 = 0
該公共弦方程為 2x - y + 1 = 0

bi) 利用 family of circle, 穿過C1和C2的方程為 :
x^2 + y^2 - 2x + 3y - 6 + k(x^2 + y^2 - 4x + 4y -7) = 0 --- (*)
因題目要求該直線穿過(1 , -2 ) , 因此代入該點到(*)
左方必定會等於右方 , 因此可得
1^2 + (-2)^2 -2(1) + 3(-2) - 6 + k( 1^2 + (-2)^2 - 4(1) + 4(-2) -7 ) = 0
1 + 4 - 2 - 6 - 6 + k( 1 + 4 - 4 - 8 - 7) = 0
-9 - 14k = 0
k = -9/14

將k代入 (*) , x^2 + y^2 - 2x + 3y - 6 + (-9/14)(x^2 + y^2 - 4x + 4y -7) = 0
5x^2 + 5y^2 + 8x - 6y - 21 = 0
因此 , 該方程為 5x^2 + 5y^2 + 8x - 6y - 21 = 0

bii) 把(*) 重新排列 ,
(1 + k)x^2 + (1 + k)y^2 + (-2 -4k)x + (3 + 4k)y - 6 - 7k = 0
一圓形的圓心為( -D/2 , -E/2 ) , 因此(*)的圓心為 [ 1+2k , -(3+4k)/2 ]
因為該圓心穿過直線 x+y-7=0 , 所以將圓心代入直線。
1 + 2k - (3+4k)/2 -7 = 0
2 + 4k -3 - 4k - 14 = 0
-15 = 0 (錯誤)
M2 : 把(*)內的兩個圓方程互換 , 得出
(1 + k)x^2 + (1 + k)y^2 + ( -4 - 2k)x + (4 + 3k) -6k - 7 = 0
圓心 : [ (2 + k) , -(4 + 3k)/2 ]
2 + k - (4 + 3k)/2 - 7 = 0
4 + 2k - 4 - 3k - 14 = 0
-k - 14 = 0
k = -14
方程 : -13x^2 - 13y^2 + 24x -38y + 77 = 0

biii) 面積6pi , (pi)(r^2) = 6pi
所以半徑 = 開方6
圓半徑 = 開方(h^2 + k^2 - F)
(2+k)^2 + [(4+3k)/2]^2 + 6k + 7 = 6
4k^2 + 16k + 16 + 9k^2 +24k + 16 + 24k + 28 = 24
13k^2 + 64k + 36 = 0
再用解2次方程方法 , 得到兩個k的值 ,
唔好意思..因為我冇帶計數機..
得兩個k的值後再代入(*) 便可得到兩個圓方程。
面積是6pi的圓是有兩個可能性。

(a)
C1及C2的公共弦方程為
(x+y-2x+3y-6)-(x+y-4x+4y-7)=0
2x-y+1=0//
(b)
(i)設所求的圓方程為
x+y-2x+3y-6+k(x+y-4x+4y-7)=0,其中k為一實常數,且k≠0。
把(1, -2)代入S的方程,可得
(1)+(-2)-2(1)+3(-2)-6+k[(1)+(-2)-4(1)+4(-2)-7]=0
-9-14k=0
k=-9/14
所求的圓方程為
x+y-2x+3y-6-(9/14)(x+y-4x+4y-7)=0
14(x+y-2x+3y-6)-9(x+y-4x+4y-7)=0
14x2-9x2+14y2-9y2-28x+36x+42y-36y-84+63=0
5x2+5y2+8x-6y-21=0
(ii)