
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
1. The length of the common chord of the circles $(x^2+y^2=4)$ and $(x^2+y^2+2ay-6=0 (a > 0))$ is $2\sqrt{3}$. Find the value of $a$.
2. In the Cartesian coordinate plane, the number of lines that are at a distance of $\sqrt{2}$ from point $A(1,2)$ and at a distance of $2\sqrt{2}$ from point $B(4,5)$ is \_\_\_\_.
3. Given that $F_1$ and $F_2$ are the two foci of an ellipse, and $\overrightarrow{MF_1} \cdot \overrightarrow{MF_2}=0$, all points $M$ are inside the ellipse. Find the range of values for the eccentricity of the ellipse.
4. The famous mathematician Hu Luogeng once said, "The combination of shape and form is a hundred times good, and the separation of families is a thousand times bad." In fact, many algebraic problems can be transformed into geometric problems for solving. For example, $\sqrt{(x-a)^2+(y-b)^2}$ can be transformed into the distance between points $M(x,y)$ and $N(a,b)$ on the plane. Based on this perspective, find the minimum value of $f(x)=\sqrt{x^2+4x+20}+\sqrt{x^2+2x+10}$.
5. Let $M(x_1,y_1)$ and $N(x_2,y_2)$ be two distinct points, and let $l: ax+by+c=0$. Define $\delta = \frac{ax_1+by_1+c}{ax_2+by_2+c}$. The following statements are given:

$①$ No matter the value of $\delta$, point $N$ is never on line $l$;
$②$ If line $l$ is perpendicular to the midpoint of segment $MN$, then $\delta=1$;
$③$ If $\delta=-1$, then line $l$ passes through the midpoint of segment $MN$;
$④$ If $\delta > 1$, then points $M$ and $N$ are on the same side of line $l$, and $l$ intersects the extension of segment $MN$.

Find the sequence number(s) of the correct statement(s).

Present the answer in LaTex format: \boxed{Your answer}
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