solve the system:
y^2=1-x^2
y^2=x^2-3x+2
since you know y^2 in terms of x, you can substitute either of them to the other equation,, in this case, i will substitute the first equation into the second, getting:
1-x^2=x^2-3x+2
then combine similar terms to one side of the equation, yielding:
0=2x^2-3x+1
since it is factorable, there shouldn't be any problem,, getting the factor:
0=(2x-1)(x-1)
x=1/2 or x=1
substituting this back to our original equation, (note: you can substitute x values to either equation, they will yield same answer anyway)
y^2=1-x^2 ; if x=1/2,
y=plus or minus squareroot of 3/4
and if x=1,
y=0
therefore, points of intersection are:
(1/2,positive sqrt of 3/4)
(1/2,negative sqrt of 3/4)
(1,0)
=)