圆的切线方程公式证明过圆(x-a)^2+(y-b)^2=r^2上点P(x0,y0)的切线方程为(x0-a)(x-a)+(

1、圆心（a,b）和切点(x0,y0)的斜率为(y0-b)/(x0-a)所以切线的斜率为-(x0-a)/(y0-b)因为切线过（x0,y0）所以切线为y=-(x0-a)/(y0-b)(x-x0)+y0整理得(x0-a)(x-x0)+(yo-b)(y-yo)=0① 因为(x0-a)^2+(y0-b)^2=r^2②①②两式相加得到(x0-a)(x-a)+(yo-b)(y-b)=r^22、可知圆心为（-D/2,-E/2）代入①式得到(x0+D/2)(x-x0)+(yo+E/2)(y-y0)=0③因为x0^2+y0^2+Dx0+Ey0+F=0④把③④相加得到x0x+y0y+D[(x+x0)/2]+E[(x0+x)/2]+F=0(问题是错误的,图片问题是正确的)3、x2+y2+Dx+Ey+F=0(x+D/2)^2+(y+E/2)^2=D^2/4+E^2/4-f所以圆心O(-D/2,-E/2),r^2=D^2/4+E^2/4-F设A(x0,y0) 切点是BAO^2=(x0+D/2)^2+(y0+E/2)^2OB^2=r^2=D^2/4+E^2/4-fOAB是直角三角形所以AB^2=OA^2-OB^2=(x0+D/2)^2+(y0+E/2)^2-D^2/4-E^2/4+F=x0^2+Dx0+y0^2+Ey0+F所以切线AB长=√(x0^2+Dx0+y0^2+Ey0+F)用勾股定理显然可得AB长=√[(x0-A)^2+(y0-B)^2-r^2]