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求经过坐标原点和点P(1,1),并且圆心在直线2x+3y+1=0上的圆的方程

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问题详情:

求经过坐标原点和点P(1,1),并且圆心在直线2x+3y+1=0上的圆的方程____________.

【回答】

 (x-4)2+(y+3)2=25.

解析:设圆心为C(xy).显然,所求圆的圆心在OP的垂直平分线上,OP的垂直平分线方程为求经过坐标原点和点P(1,1),并且圆心在直线2x+3y+1=0上的圆的方程求经过坐标原点和点P(1,1),并且圆心在直线2x+3y+1=0上的圆的方程 第2张,即xy-1=0.

解方程组求经过坐标原点和点P(1,1),并且圆心在直线2x+3y+1=0上的圆的方程 第3张得圆心C的坐标为(4,-3).                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                         

又圆的半径r=|OC|=5,

∴所求圆的方程为(x-4)2+(y+3)2=25.

知识点:圆与方程

题型:填空题

标签:2x 原点 3y 圆心 p1