Solve the system

x^2+y^2=4

x+y=2

This is the intersecction of a line  and a circle

Using the second equation we can write y = 2 -  x

And substituting this into the first equation for y, we have

x^2 + (2 - x)^2 = 4   simplify

x^2 + x^2 - 4x + 4 = 4   subtract 4 from both sides  and simplify

2x^2 - 4x = 0  factor

2x(x - 2) = 0       setting each factor to 0, we have that x = 0, and x = 2

And using x + y = 2

When x = 0, y = 2

When x = 2, y =0  so the answers are (0,2) (2,0)

Here's a graph......https://www.desmos.com/calculator/eakj6g1q0t

$$\\x^2+y^2=4\qquad(1)\\\\
x+y=2 \;\;\rightarrow \;\;y=2-x\qquad(2)\\\\
sub\;\;2\;into\;1\\\\
x^2+(2-x)^2=4\\\\
x^2+4+x^2-4x=4\\\\
2x^2-4x=0\\\\
2(x-2)(x+2)=0\\\\
x=\pm 2$$

You can find the y values :)

x = 2 implying y = 0 is ok. but x = -2 leads to a contradiction.

Also, the equations are symmetric in x and y, so it's possible to switch the x and y values to obtain a second solution corresponding to x=2, y=0.

An alternative method of solution would be to square the second equation and then combine that with the first leading to 2xy = 0, from which either x = 0, or y = 0.

Then substitute into the second equation, (not the first), to find the corresponding y or x value.

Thanks guys,

sometimes it pays to finish a question. 

I suppose it also pays to look at a question sensibly before plowing headlong into the algebra   :)