Question

已知圆x2+y2+x-6y+m=0和直线x+2y-3=0交于P、Q两点，若OP⊥OQ(O是原点)，求m的值.

Answer 1

x+2y-3=0, x = 3-2y
代入x²+y²+x-6y+m=0， 5y² - 20y + 12+m = 0
△=400-4*5*(12+m) = 20(8-m)
y1 = (20 + √△)/10 = 2 + (√△)/10, x1 = 3 - 2(2 + √△/10) = -1 -(√△)/5
y2 = (20 - √△)/10 = 2 - (√△)/10, x2 = 3 - 2(2 - √△/10) = -1 +(√△)/5
P(-1 -(√△)/5, 2 + (√△)/10)
Q(-1 +(√△)/5, 2 - (√△)/10)
OP的斜率k1 = [2 + (√△)/10]/[-1 -(√△)/5]
OQ的斜率k2 = [2 - (√△)/10]/[-1 +(√△)/5]
OP垂直于OQ， k1*k2= -1
{[2 + (√△)/10]/[-1 -(√△)/5] }* {[2 - (√△)/10]/[-1 +(√△)/5]} = -1
(4 - △/100)/(1 - △/25)= -1
△=100
20(8-m)=100
m=3

Answer 2

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追问

斜率不存在呢？