Find the points of intersection between circle and a parable:
circle: $x^2 + y^2 - 2x + 4y - 11 = 0$
parable: $y = (-x^2+ 2x + 1 - 2\sqrt{3})$
I don't understand how to solve this, I really tried, but was useless.
If anyone can help me, I am very grateful.
The point at which the circle and parabola interect, $(x_0,y_0),$ satisfies both equations, so
$x_0^2+y_0^2-2x_0+4y_0-11=0 \tag{1}$
$\color{green}{y_0=-x_0^2+2x_0+1-2\sqrt{3}} \tag{2}$
Now, solve $(1)$ and $(2)$ simultaneously!
Hint: substitute $\color{green}{y_0=-x_0^2+2x_0+1-2\sqrt{3}}$ into $(1)$ to give:
$$x_0^2+[\color{green}{-x_0^2+2x_0+1-2\sqrt{3}}]^2-2x_0+4[\color{green}{-x_0^2+2x_0+1-2\sqrt{3}}]-11=0 \tag{3}.$$
Solve $(3)$, for $x_0$.
Once you have solved $(3)$ for $x_0$, sub this into either $(1)$ or $(2)$ to find $y_0$.