Set of points that satisfy the sum of the distances between a point P and points A(-2,0) and B(2,0) = 7 So, I have to find the set of points whose distance to point $A(-2, 0)$ plus the distance to point $B(2, 0)$ is equal to $7$.
I am solving it this way:
$ d(P, A) + d(P, B) = 7  \Leftrightarrow $
$ \sqrt{(x+2)^2+y^2}+\sqrt{(x-2)^2+y^2}=7 \Leftrightarrow $
$ (x+2)^2+y^2+(x-2)^2+y^2=49 \Leftrightarrow $
$ x^2+4x+4+y^2+x^2-4x+4+y^2=49 \Leftrightarrow $
$ 2x^2+2y^2=49-8 \Leftrightarrow $
$x^2+y^2=\frac{41}{2} $
So, we have a circle centered in the Origin and radius $\sqrt{\frac{41}{2}}$
Now, when I drew this circle, and measured the distance from any point belonging to the circle to A adding to the distance of that same point to B, I get the result of around $9$
And effectively, $2\sqrt{\frac{41}{2}} \approx 9.05$
Where am I going wrong?

When you expand $(a+b)^2$, you get $\color{blue}{\text{cross terms}}$ also as
$$(a+b)^2=a^2+b^2+\color{blue}{2ab}$$
You have missed
$$(x+2)^2+y^2+(x-2)^2+y^2+\color{blue}{2\cdot \sqrt{(x+2)^2+y^2}\cdot\sqrt{(x-2)^2+y^2}}=49$$
One more squaring will clear the square roots and finally give correct equation (of an ellipse).
If this is tedious, one can recognize the curve with given property of constant sum of distances from two points as an ellipse. And try to find its parameters.
The ellipse will be centered at midpoint of two foci, $(0,0)$. Major axis $2a=7 \Rightarrow a^2=49/4$. Distance of center from foci, $ae=2$. Minor axis is $2b$ where $b^2=a^2-a^2e^2=33/4$. Therefore required ellipse is
$$\frac{4x^2}{49}+\frac{4y^2}{33}=1$$
