x^2+y^2-2tx-2 t^2 y+4t
=(x^2-2tx)+(y^2-2 t^2 y)+4t
=(x^2-2tx+t^2)+(y^2-2 t^2 y+t^4)-(t^4+t^2-4t)
=(x-t)^2+(y-t^2)^2-[(t^4+t^2+4t^2)-(4t^2+4t)]
=(x-t)^2+(y-t^2)^2-[(t^2+t)^2-4t(t+1)]
=(x-t)^2+(y-t^2)^2-[t^2(t+1)^2-4t(t+1)]
=(x-t)^2+(y-t^2)^2-[t(t+1)-/2]^2+1
因为x^2+y^2-2tx-2 t^2 y+4t=0
所以(x-t)^2+(y-t^2)^2-[t(t+1)-/2]^2+1=0
而(x-t)^2和(y-t^2)^2均不小于0,
所以(x-t)^2=(y-t^2)^2=0且-[t(t+1)-/2]^2+1=0(*)
所以圆C:x^2+y^2-2tx-2 t^2 y+4t=0必过点(t,t^2)
而t的值解(*)式可得