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.