已知圆X^2+y^2+dx+ey+f=0.证明以p(xo,yo)为切点的切线方程是xxo+yyo+d*x+xo/2+ey

设Q(x,y)在以P(x0,y0)为切点的切线上,因为x^2 + y^2 +dx + ey + f =0的圆心为M(-d/2,-e/2),所以
MP⊥QP,所以直线MP与QP的斜率的乘积为-1,即(y-y0)/(x-x0) * (y0 + e/2)/(x0 + d/2) = -1,展开可得xx0 + yy0 + d(x+x0)/2 + e(y+y0)/2 + f = (x0)^2 + (y0)^2 + dx0 + ey0 + f =0,其中
(x0)^2 + (y0)^2 + dx0 + ey0 + f =0是因为点P(x0,y0)在圆周上.