园x^2 +y^2 +x-6y+3=0上的两点P,Q,关于直线kx-y+4=0对称.OP垂直OQ,求直线PQ的方程
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园x^2 +y^2 +x-6y+3=0上的两点P,Q,关于直线kx-y+4=0对称.OP垂直OQ,求直线PQ的方程
园x^2 +y^2 +x-6y+3=0上的两点P,Q,关于直线kx-y+4=0对称.OP垂直OQ,求直线PQ的方程
园x^2 +y^2 +x-6y+3=0上的两点P,Q,关于直线kx-y+4=0对称.OP垂直OQ,求直线PQ的方程
x^2 +y^2 +x-6y+3=0
(x+0.5)²+(y-3)²=6.25
所以:O坐标:(-0.5,3) ,圆半径为2.5
直线kx-y+4=0必过圆心可解得:k=2
所以直线PQ的斜率为-0.5
设直线PQ方程为:-0.5x-y+b=0
可求得直线PQ与直线kx-y+4=0的交点坐标为:((-8-4b)/5,(4-4b)/5)
交点与圆心距离=半径/根号2 {几何知识}
即:
(-0.5+(8+4b)/5)²+(3-(4-4b)/5)²=(2.5/根号2)²
这样可解得两个b的值.
代入:-0.5x-y+b=0即可得直线PQ的两个方程.
此直线必过圆心
可解出k
之后就好办了
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