已知圆x^2+y^2+x-6y+m=0和直线x+2y-3=0交于p、q两点，且op垂直于oq,o为坐标原点，求圆的圆心坐标及半径长

设两交点坐标P(x1,y1),Q(x2,y2)
直线方程 x=3-2y，代入圆方程化简得：
5y^2-20y+(12+m) = 0

P,Q在圆上，因此
y1+y2 = -b/a = 4
y1y2 = c/a = (12+m)/5

直线PO斜率 k1 = y1/x1
直线OP斜率 k2 = y2/x2
PO⊥OQ
====> k1k2 = -1
======> x1x2 + y1y2 = 0

P，Q在直线上
(3-2y1)(3-2y2) + y1y2 =0 ,
5y1y2 - 6(y1+y2) + 9 = 0
把y1+y2和y1*y2的值代入，得
5*(12+m)/5 - 6*4 + 9 = 0

====> m = 3
圆 x^2+y^2+x-6y+3=0
(x+1/2)^2+(y-3)^2=(5/2)^

圆心(-1/2,3).半径5/2