Problem 7

Consider the curves x2 + y2 = 1 and 2x2 + 2xy + y2–2x–2y = 0. These curves intersect at two points, one of which is (1, 0). Find the other one.
Solution.
figure fig_Problem_7.jpg
We have 2x2 + 2xy + y2 − 2x − 2y = 0, then (2x − 2)(x + y) + y2 = 0.
Since x2 + y2 = 1 is the other curve, we have y2 = 1 − x2 and this gives to us
2(x − 1)(x + y) + (1 − x2) = 0
2(x − 1)(x + y) − (x − 1)(x + 1) = 0
(x − 1)(2y + x − 1) = 0.
Then x = 1 or y = (1 − x)/(2).
The case x = 1 leads to the point (1, 0).
In the other case we have the system of equations
y = (1 − x)/(2) x2 + y2 = 1
with the solutions (1, 0) and ( − 3)/(5),  (4)/(5).