Question

过点P（-8,0）,引圆C:X^2+Y^2-2X+10Y+4=0的割线,求被此圆载得的弦长的中点轨迹方程

Answer 1

圆C:x^2+y^2-2x+10y+4=0,(x-1)^2+(y+5)^2=22

C(1,-5)

弦AB的中点P(x,y)

k(AB)*k(CP)=-1

[y/(x+8)]*[(y+5)/(x-1)]=-1

弦的中点的轨迹方程是园：(x+2.5)^2+(y+3.5)^2=26.5