Consider the curves *x*^{2} + *y*^{2} = 1 and 2*x*^{2} + 2*xy* + *y*^{2}–2*x*–2*y* = 0. These curves intersect at two points, one of which is (1, 0). Find the other one.

We have 2*x*^{2} + 2*xy* + *y*^{2} − 2*x* − 2*y* = 0, then (2*x* − 2)(*x* + *y*) + *y*^{2} = 0.

Since *x*^{2} + *y*^{2} = 1 is the other curve, we have *y*^{2} = 1 − *x*^{2} and this gives to us*x* = 1 or *y* = (1 − *x*)/(2).

2(*x* − 1)(*x* + *y*) + (1 − *x*^{2}) = 0

2(*x* − 1)(*x* + *y*) − (*x* − 1)(*x* + 1) = 0

(*x* − 1)(2*y* + *x* − 1) = 0.

Then
The case *x* = 1 leads to the point (1, 0).

In the other case we have the system of equations
*y* = (1 − *x*)/(2)
*x*^{2} + *y*^{2} = 1
with the solutions (1, 0) and ⎛⎝( − 3)/(5), (4)/(5)⎞⎠.