<s>[INST] Finding the points where a circle intersects an axis A circle has the equation:

x²+y²+4x-2y-11 = 0

What would be the coordinates of the points where the circle intersects with the y-axis and how would you calculate it?
 [/INST] $x^2+y^2+4x-2y-11=0$
The points of intersection with the $y$ axis are when $x=0$. Thus, plug in $x=0$ and solve.
$$0^2+y^2+4(0)-2y-11=0$$
$$y^2-2y-11=0$$
Use the quadratic formula: $\large\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ where $a=1, b=-2, c=-11$
$$\frac{-(-2)\pm \sqrt{4-(-44)}}{2}$$
$$\frac{2\pm \sqrt{48}}{2}$$
$\sqrt{48}=\sqrt{8\cdot 6}=\sqrt{4\cdot 2\cdot 3\cdot 2}=4\sqrt{3}$
$$\frac{2\pm 4\sqrt{3}}{2}=1\pm 2\sqrt{3}$$
So, your coordinates for intersection with the $y$-axis are: $(0,1+ 2\sqrt{3})$ and $(0,1- 2\sqrt{3})$
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