Y=a-x, substitute x? +y？ =4, so:

x？ +(a-x)？ =4

Finishing, get: 2x? -2ax+a？ -4=0

Discriminant△ = (-2a)? -4 2 (a？ -4)=-4a？ +32

Make △ > 0, get: -4a? +32 >0

Answer? & lt eight

-2√2 & lt； a & lt2√2

Let △=0 and the solution is: a=2√2 or a=-2√2.

Manufacture △ <

Classification discussion:

( 1)、-2√2 & lt； a & lt2√2，△& gt； 0

These equations have two sets of solutions.

(2), a=2√2 or a=-2√2, δ= 0.

These equations have a set of solutions.

(3), a & lt-2√2 or a>2 √ 2, delta < 0.

These equations have no solution.