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OP垂直于OQ，求圆心及坐标

Answer 1

解起来很复杂的样子。。。。。
(x+1/2)²+(y-3)²=-m+1/4+9=37/4-m圆心（-1/2,3）圆心到直线距离=|-1/2+6-3|/√1+2²=√5/2
设P(3-2a,a),Q(3-2b,b)由于OP垂直于OQ所以a/(3-2a) * b/(3-2b) = -1
所以 ab = -9+6a+6b-4ab 即 5ab-6(a+b)+9 =0 (0)
因为P、Q都在圆上所以 将P带入圆方程为 5a²-20a +12-m =0 (1)
a、b为 方程(1)的两个解 所以 ab= (12-m)/5;a+b =4 带入方程(0)
得到 12-m -24 +9 =0 则m= -3
m带入方程(1)得到 a为3 和1 则P(1,1) Q(-3,3)

Answer 2

x^2+y^2+x-6y+m=0

(x+1/2)²+(y-3)²=-m+1/4+9=37/4-m
圆心（-1/2,3）
圆心到直线距离=|-1/2+6-3|/√1+2²=√5/2
所以
半径=√5/2*√2=√10/2
所以
37/4-m=5/2

m=37/4-10/4=27/4

Answer 3

x^2+y^2+x-6y+m=0 (1)
(x+1/2)^2+(y-3)^2 - 25/6+m=0
centre O( -1/2 , 3)
let P(x1,y1), Q(x2,y2)
OP垂直于OQ
=> (y1-3)/(x1+1/2) = -(x2+2)/(y2-3)
y1y2-2(y1+y2)+9 = -(x1x2+2(x1+x2)+4) (2)

x+2y-3=0 (3)
sub (2) into (1)
(3-2y)^2+y^2+ (3-2y)+m=0
5y^2-14y+12+m=0 (8)
y1+y2 = 14/5 (4)
y1y2=(12+m)/5 (5)
Similarly sub (2) into (1)
x^2+[(3-x)/2]^2+x-6[(3-x)/2]+m=0
6x^2+(3-x)^2+4x-12(3-x)+4m=0
7x^2+10x+4m-27=0
x1+x2= -10/7 (6)
x1x2=(4m-27)/7 (7)

sub(4)(5)(6)(7) into (2)
y1y2-2(y1+y2)+9 = -(x1x2+2(x1+x2)+4)
(12+m)/5 -2(14/5) +9 = -((4m-27)/7+2(-10/7) +4 )
(29+m)/5= -(4m-19)/7
203+7m=-20m+95
27m=-108
m=-4

from (8)
5y^2-14y+12+m=0
5y^2-14y+8=0
(y-2)(5y-4)=0
y=2 or 4/5
y=2, x=-1
y=4/5, x= 7/5
P,Q = (-1,2) , (7/5,4/5)