已知直角三角形ABC的斜边AB,且A(-5,1),B(3-2),求顶点C的轨迹方程

AB^2=(1+2)^2+(-5-3)^2=73
设C坐标是（x,y)
AC^2=(x+5)^2+(y-1)^2
BC^2=(x-3)^2+(y+2)^2
所以有：(x+5)^2+(y-1)^2+(x-3)^2+(y+2)^2=73
化简后即得轨迹方程.