已知圆C1:X²+y²-10x-6y+32=0,动圆C2:x²+y²-2ax-2(8-a)y+4a+12=0(a∈R)

求证：圆C1与圆C2相交于两个定点

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C2-C1，得皮灶：燃物扮10x+6y-32-2ax-(16-2a)y+4a+12=0
(10-2a)x-(10-2a)y+4a-20=0
(10-2a)(x-y-2)=0
y=x-2
代入C1：x^2+x^2-4x+4-10x-6x+12+32=0
2x^2-20x+48=0
x^2-10x+24=0
x=4或6
y=2或4
所以圆C1与圆C2相交于两个蚂租定点(4,2)和(6,4)

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C1:(x-5)^2+(y-3)^2=2
C1(5,3) r1=√2

C2:(x-a)^2+(y-8+a)^2=a^2+(a-8)^2-4a-12=2a^2-20a+52=2(a-5)^2+2
C2(a,8-a) r2=√[2(a-5)^2+2]

C1到C2距离d^2=(a-5)^2+(5-a)^2=2(a-5)^2
d=√2|a-5|
r2-r1<颤型d圆C1、茄高猜圆念含C2相交，2个交点

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