Question

已知直线x-y+2=0和圆x2+y2+2x-6y+m=0交于P、Q两点,OP垂直于OQ(其中O为原点)(1)求m

Answer 1

x²+y²+2x-6y+m=0
(x+1)²+(y-3)²=10-m
圆心（-1，3）半径=√(10-m)
x-y+2=0
y=x+2
代入圆的方程
x²+(x+2)²+2x-6(x+2)+m=0
2x²+6x+4-6x-12+m=0
2x²+m-8=0
x1+x2=0
x1x2=(m-8)/2
根据题意
Kop×Koq=-1
(y1-0)/(x1-0)×(y2-0)/(x2-0)=-1
y1y2+x1x2=0
(x1+2)(X2+2)+x1x2=0
x1x2+2(X1+x2)+4+x1x2=0
x1x2+2=0
(m-8)/2+2=0
m-8=-4
m=4
还有问题，请补充

Answer 2

椭圆：x²/30+y²/24=1