Question

已知圆X^2+Y^2+X-6Y+M=0和直线X+2Y-3=0交于P.Q两点,且向量OP·OQ=0（O为坐标原点,求该圆的

Answer 1

设P（x1,y1）Q（x2,y2）向量OP·OQ=0,所以x1x2+y1y2=0,因为X^2+Y^2+X-6Y+M=0且X+2Y-3=0,所以两方程连列,消去x得5y^2-20y+12+M=0,所以y1y2=(12+M)/5,y1+y2=4

x1x2=(3-2y1)(3-2y2)=9-6y1-6y2+4y1y2=9-6(y1+y2)+4y1y2=9-24+4(12+M)/5=4(12+M)/5-15

x1x2+y1y2=4(12+M)/5-15+(12+M)/5=0,所以M=3,X^2+Y^2+X-6Y+3=(X+1/2)^2+(Y-3)^2=25/4

所以圆心坐标（-1/2,3）,R=5/2