Question

已知曲线C：x²+y²-4mx+2my+20m-20=0.（1）求证：不论m取何实数,曲线C恒过一定

Answer 1

x^2+y^2-4mx+2my+20m-20=0

(x-2m)^2+(y+m)^2=5m^2-20m+20

(x-2m)^2+(y+m)^2=5(m-2)^2

(4-2m)^2+(-2+m)^2=16-16m+4m^2+4-4m+m^2=20-20m-5m^2=5(m-2)^2

恒过(4,-2)点