急需答案，謝謝了

x=2+3cosθ (1)
y=-1+3sinθ (2)
from (1) and (2)
[(x-2)/3]^2 +[(y+1)/3]^2 =1 (3)
x-y=0 (4)
sub (4) into (3)
[(x-2)/3]^2 +[(y+1)/3]^2 =1
[(x-2)/3]^2 +[(x+1)/3]^2 =1
(x-2)^2 +(x+1)^2 =9
2x^2-2x+5 =9
2x^2-2x-4 =0
x^2-x-2 =0
(x-2)(x+1) =0
x=2 or -1
交點(diǎn) =(2,2) or (-1,-1)