✔ 最佳答案
C1: x^2 + y^2 - 4 = 0. Centre (0,0) radius=2
C2: x^2 + y^2 -10y + 24 =0. Centre (0,5) radius=1
Let the centre of the required circle (h,k). radius=a
Then √(h^2+k^2)=a+2...(1) √[(h^2)+(k-5)^2)]=a+1...(2)
So √[(h^2+k^2)]-2=√[(h^2)+(k-5)^2)]-1...(3)
√[(h^2+k^2)]=√[(h^2)+(k-5)^2)]+1
[(h^2+k^2)]=[(h^2)+(k-5)^2)]+2√[(h^2)+(k-5)^2)]+1
0=(25-10k)+2√[(h^2)+(k-5)^2)]+1
√[(h^2)+(k-5)^2)]=5k-13
(h^2)+(k-5)^2=25k^2-130k+169
h^2+k^2-10k+25=25k^2-130k+169
h^2-24k^2+120k-144=0
So the equation of the locus of the centre of the circle C is
x^2-24y^2+120y-144=0