x0+y0=2a- 1 - ( 1)
x0^2+y0^2=a^2+2a-3 - (2)
(1) minus (2).
2x0y0=(2a- 1)？ -(a? +2a-3)=(4a？ -4a+ 1)-(a？ +2a-3)=3a？ -6a+4
x0y0=(3/2)a？ -3a+2
After the formula
x0y0=(3/2)(a- 1)？ + 1/2
Because the radius of a circle cannot be less than 0.
So there is an A+2A-3 >: 0
Get an a
Because a straight line intersects a circle, the distance from the center of the circle (0,0) to the straight line is less than or equal to the radius, so there is inequality.
(2a- 1)？ /( 1+ 1)& lt； =a？ +2a-3
4a？ -4a+ 1 & lt； =2a？ +4a-6
2a？ -8a+7 & lt； =0
2- (radical 2/2)= 2
The comprehensive value range of a is [2- (root number 2/2), 2+ (root number 2)/2].
So when a=2- (root number 2)/2, x0y0 has a minimum value.