x^2+y^2 = 4 is the eqn of the circle with center @ origin (0,0) and radius = 4

y = -x + 4 is a line with y-intercept of 4 and a -ve slope of -1.
The line intercepts the x-axis at (4,0)

Hence the answer is 2

x^2+y^2=4,
x+y=4.

I think you know the pattern of the first equation.
It is cercle with the center at (0;0), and radius 4.

The second one is the line, y=4-x, if y=0, then x=4, and if x=0, then y=4. Then we have 2 intersect points, hence the answer would be indeed 2.
Well, i tried to solve this one algebraically, but i got discriminant that is less then 0... Literally it means those eqs. don't have common roots Then i thought it would be less time consuming to solve this one graphically....)

GMATV09, hehe)))

gmatv09 and NikolayZ -
Isn't x^2+y^2 = 4 a circle with radius 2 ( Not 4)and center at (0,0) ??

ooops, you are right Ayushr. My school math is 7 years away so i made this silly mistake. indeed the radius must be 2.
Then the answer to this particular problem is 0.
If the circle equation looks like : x^2+y^2=16(radius 4), then there are 2 common roots.

P.s. Then it is clear why the discriminant is negative and these line and circle have no common roots.

Guys OA is 2. We are missing something. This is a GMAT clubs problem.