在极坐标系中,直线psin(Θ+派/4)= 2被圆p=4截得玄长是多少?

PSIN(θ+π/4)=P(SINθ*COSπ/4+COSθ*SINπ/4)=√2/2*PSINθ+√2/2*PCOSθ
=√2/2(X+Y)=2
=>X+Y=2√2 =>X+Y-2√2=0
P=4 =>P²=16 =>X²+Y²=16 圆心(0,0),半径R=4.
圆心到直线距离:D=|-2√2|/√(1+1)=2
设弦长为L,根据垂径定理:(L/2)²=R²-D²=4²-2²=16-4=12
=>L=4√3
应此,答案为4√3.