已知园经过点A（2,-3）和B（-2,-5） 1.若圆心在直线x-2y-3=0上,求园的方程； 2.若园与直线l:2x+

已知园经过点A（2,-3）和B（-2,-5） 1.若圆心在直线x-2y-3=0上,求园的方程； 2.若园与直线l:2x+y-5√
已知园经过点A（2,-3）和B（-2,-5）
1.若圆心在直线x-2y-3=0上,求园的方程；
2.若园与直线l:2x+y-5√2+4=0相切,求圆的方程

圆心设为(x0,y0)满足x0-2y0-3=0,即为(2y0+3,y0)由于圆心到圆上任意两点距离相等,所以到A点和B点距离相等,列方程：                                  (2y0+3-2)^2+(y0+3)^2=(2y0+3+2)^2+(y0+5)^2解方程得：                                                                 y0=-2故                                                        x0=2y0+3=-1圆心为(-1,-2).计算圆的半径,也就是到A点或B点的距离,为：                                           根号[(-1-2)^2+(-2+3)^2]=根号10圆的方程为：                                                  (x+1)^2+(y+2)^2=10  圆心在与AB垂直并过AB重点的直线上（能理解吗?）计算AB方程为(计算过程略)：                                                             x-2y-8=0与AB垂直并过AB重点的直线方程(计算过程略)为：                                                            2x+y+4=0                                 (其实这条线和题中所给的圆的切线是平行的)圆心设为(x1,-2x1-4)圆心到直线的距离与到A点距离相等,列方程：             (x1-2)^2+(-2x1-4+3)^2=[2x1+(-2x1-4)-5根号2+4]^2/(2^2+1^2)                                        (这里用到了点到直线的距离公式)解方程,得：                                                             x1=±1故圆心为(1,-6)或(-1,-2)圆的方程为：                                                   (x-1)^2+(y+6)^2=10                                                                 或                                                   (x+1)^2+(y+2)^2=10 手打得很辛苦啊o(╯□╰)o 祝学习进步!