Intersection of a circle and a triangle If we have a point $(0,6)$ which is the vertex of an equilateral triangle the distance of whose all vertices are equal from origin.Draw a circle of radius 3 centered at $(0,1)$.How many number of maximum intersection of circle and triangle?
I don't know how to solve because i am week in geometry, from picture i find it may be 3 or 4,but i am not sure.Please give me some direction to solve this. 
 A: This may not be the fastest or most sophisticated method, but I "think" it works. It is an algebraic approach to solving this type of problem. A more clever person than I can attest one way or the other.
First recognize that the slope of an equilateral triangle is $\pm\sqrt{3}$. From here, we can find the equation of the line that is a side of the triangle. Either one will work as whatever result we get for the number of intersection points for one side will be equal to the other. Thus, we can just double our answer.
Equation of a line $y=mx+b$, where $m=\pm\sqrt{3}$ (we will choose $+\sqrt{3}$), and $b=6$. Thus, we get:
$$y=\sqrt{3}x+6$$
What we are ultimately looking for is the number of intersection points between our line and the circle. The equation of a cicle is $(x-a)^2+(y-b)^2=r^2$. Where we have $a=0$, $b=1$ and $r=3$. Giving us:
$$x^2+(y-1)^2=9$$
Substitute, $y=\sqrt{3}x+6$ into $x^2+(y-1)^2=9$. After doing a little algebra we get:
$$4x^2+10\sqrt{3}x+16=0$$
Which is a quadratic with $a=4$, $b=10\sqrt{3}$ and $c=16$.
Using the discriminant of the quadratic:
If $b^2-4ac<0$, there are no points of intersection.
If $b^2-4ac=0$, the line is tangent to the circle and therefore has one point of intersection.
If $b^2-4ac\gt 0$, there are two points of intersection.
$$(10\sqrt{3})^2-4(4)(16)=44$$
Thus, each line intersects the circle at two points giving us $4$ total points of intersection.
