I'm deriving the standard equation for an ellipse from two arbitrary points \(F_1\) and \(F_2\).
An ellipse around these two *focal points* is a set of points \(P\) for which the sum of the distances to \(F_1\) and \(F_2\) is a constant \(s\), where \(s\) may be any value greater \(\overline{F_1 F_2}\).
$$ \overline{F_1 P} + \overline{F_2 P} = s $$
Let's sketch this out with a little SVG (luckily, SVG has an ellipse element).

We have our focal points \(F_1 = (-c, 0)\) and \(F_2 = (c, 0)\).
The midpoint of \(\overline{F_1 F_2}\) is the point of origin and center of the ellipse \(O = (0,0)\).
The points \(V_1 = (a, 0)\) and \(V_2 = (0, b)\) are the *vertices* of the ellipse.
Parameter \(a\) is the *semi-major axis* (maximal radius of the ellipse), \(b\) the *semi-minor axis* (minimal radius of the ellipse), and \(c\) the *linear eccentricity* (distance from the center to any of the two foci).

The sum of the distances from \(F_1\) and \(F_2\) to \(V_1\) is \(2a\), so in general \(s = 2a\). $$ \overline{F_1 V_1} + \overline{F_2 V_1} = (c + a) + (a - c) = 2a $$

In case you don't believe me, consider the points \(P\), \(F_1\) and \(Q = (-c, y)\). If \(-c = x\), then the distance obviously is \(\sqrt{(x+c)^2 + y^2} = y\). Otherwise, those three points form a right triangle with the hypotenuse \(\overline{P F_1}\). Line segment \(\overline{P Q}\) is parallel to the x-axis and has length \(x + c\). Line segment \(\overline{F_1 Q}\) is parallel to the y-axis and has length \(y\). The Pythagorean theorem says \(\overline{P F_1}^2 = (x + c)^2 + y^2 \).

Now we can express the definition of the ellipse with the following equation: $$\sqrt{(x-c)^2 + y^2} + \sqrt{(x+c)^2 + y^2} = 2a .$$

At this point we use the Pythagorean theorem to substitute \(c^2 = a^2 - b^2\). $$ \begin{align*} \frac{(a^2 - b^2) x^2}{a^2} + \cancel{a^2} &= x^2 + \cancel{a^2} - b^2 + y^2 \\ \cancel{a^2 x^2} - b^2 x^2 &= \cancel{a^2 x^2} - a^2 b^2 + a^2 y^2\\ -x^2 &= -a^2 + \frac{a^2 y^2}{b^2} \\ \frac{x^2}{a^2} &= 1 - \frac{y^2}{b^2} \end{align*} $$

If \(F_1 = F_2\) (there is only one focal point) and thus \(a = b = r\), we have the special case of a circle with radius \(r\) around the center \(O\). $$ x^2 + y^2 = r^2 $$