✔ 最佳答案
１. C1: x²＋y² - 14x - 2y - 40 = 0 , C2 : x² + y² - 2x - 6y = 0
C1: x²＋y² - 14x - 2y - 40 = 0
(x-7)^2 + (y-1)^2 = 90
圓心(7,1) 半徑√90
C2: : x² + y² - 2x - 6y = 0
(x-1)^2+(y-3)^2=10
圓心(1,3) 半徑√10
兩個圓心距離=√(6^2+2^2)=√40
所以C1半徑-C2半徑=2√10=兩個圓心距離
C2內切C1
由C1: x²＋y² = 14x + 2y + 40
代入C2
14x + 2y + 40 - 2x - 6y = 0
12x-4y+40=0
3x-y+10=0
y=3x+10
代入C1: x²＋y² - 14x - 2y - 40 = 0
x^2+9x^2+60x+100-14x-6x-20-40=0
10x^2+40x+40=0
x^2+4x+4=0
(x+2)^2=0
x=-2,y=4
切點的坐標(-2,4)
公切線的方程:C1-C2
x²＋y² - 14x - 2y - 40 - (x² + y² - 2x - 6y) = 0
-12x+4y-40=0
3x-y+10=0
2
設該圓方程為
x² + y² - 25 + k(x² + y² - 4x - 6y - 3) = 0
(1+k)x^2+(1+k)y^2-4kx-6ky-(25+3k)=0
代A(8,5)
(1+k)(64)+(1+k)(25)-32k-30k-(25+3k)=0
89(1+k)-62k-25-3k=0
24k=-64
k=-8/3
圓方程為
(1-8/3)x^2+(1-8/3)y^2+(32/3)x+(48/3)y-(25-8)=0
(-5/3)x^2+(-5/3)y^2+(32/3)x+(48/3)y-(51/3)=0
5x^2+5y^2-32x-48y+51=0
