Question

a为何值时，两圆x^2+y^2-2ax+4y+a^2-5=0和x^2+y^2-2x-2ay+a^2-3=0 外切？

Answer 1

x²+y²-2ax+4y+a²-5=0
(x - a)² + (y + 2)² = 9, 圆心A(a, -2), 半径R = 3
x²+y²-2x-2ay+a²-3=0

(x -1)² + (y -a)² = 4, 圆心B(1, a), 半径r = 2
二者外切, AB = R+r = 5
AB² = 25 = (a - 1)² + (-2 - a)² = 2a² + 6a + 5
a² + 3a - 10 = 0
(a + 5)(a - 2) = 0
a = -5或a = 2

Answer 2

x²+y²-2ax+4y+a²-5=0 x²+y²-2x-2ay+a²-3=0
x²-2ax+a²+y²+4y+4=3² x²-2x+1+y²-2ay+a²=2²
(x-a)²+(y+2)²=3² (x-1)²+(y-a)=2²
(a-1)²+(-2-a)²=(2+3)²
a²-2a+1+a²+4a+4=25
a²+a-10=0
a=(-1±√41)/2