Question

圆x^2+y^2+2x+4y-3=0上到直线x+y+1=0的距离根号2的点的坐标

Answer 1

圆x^2+y^2+2x+4y-3=0

(x+1)^2+(y+2)^2=8

设x=-1+2√2cosa

y=-2+2√2sina

到直线x+y+1=0的距离

d=|-2+2√2cosa+2√2sina|/√2

=|4sin(a+π/4)-2|/√2=√2

4sin(a+π/4)-2=2或 4sin(a+π/4)-2=-2

sin(a+π/4)=1 或 4sin(a+π/4)=0

a=π/4 或 a=3π/4

点的坐标为

(1,0)或(-3,0)