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If the ellipse has radii $a=5$, $b=2$, and center $(0,0)$, how can I find the distance from the center to any point $P(x,y)$?

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    This distance will depend on the point and can vary between 2 and 5. You calculate it just like you calculate the [distance between any two points](https://en.wikipedia.org/wiki/Euclidean_distance). The ellipse is irrelevant. – 2017-01-19

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An ellipse has parametric equations: $$x=a\cos{\theta} \tag{1.1}$$ $$y=b\sin{\theta} \tag{1.2}$$ This will give you a locus of values of $x$ and $y$ which will satisfy your ellipse's equation.

You can simply use Pythagoras' Theorem. I put them in both Cartesian and Parametric form for the case of your ellipse:

$$\sqrt{x^2+y^2}=\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}}$$

As others have noted, the ellipse is irrelevant, as long as you are certain that the point $P(x,y)$ lies on the ellipse.

To check this, check if the value you evaluate for $\theta$ on equation $(1.1)$ is consistent with $(1.2)$ by substituting your values for $a,b,x,y$. If the two values of $\theta$ you evaluate are equal, it lies on the ellipse's curve. Otherwise, it does not.

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    +1 because this answer went beyond the trivial and put the distance in terms of the point being on the ellipse (which the OP didn't seem to realize was not relevant to the given x,y coordinates). – 2017-01-19
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It depends very much on how you specify the point $P$. In any case, the distance from any point of the plane $P(x,y)$ to the origin is $\sqrt{x^2+y^2}$.

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    I'm aware that this answer may be not very enlightning. If you specify what is the precise role of the ellipse in the problem, I could improve my answer. – 2017-01-19
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The equation of the ellipse is $\frac {x^2}4 + \frac {y^2}{25} = 1$ or $25x^2 + 4 y^2 = 100$ or $x^2 = 4(1 - \frac {y^2}{25})$ or $y^2 = 25(1 - \frac{x^2}4)$.

The equation for the distance between $(x,y)$ and $(0,0)$ is distance =$\sqrt{x^2 + y^2}$

So calculating the distance with only the $x$ or only the $y$ value we know:

distance =$\sqrt{x^2 + y^2} = \sqrt{x^2 + 25(1-\frac {x^2}4)} = \sqrt{25-\frac {21x^2}4}$

or

distance =$\sqrt{x^2 + y^2} = \sqrt{y^2 + 4(1-\frac {y^2}{25})} = \sqrt{4+\frac {21y^2}{25}}$

(Note: for a give $x$ or $y$ there are generally 2 possible $y$s or $x$s.)

Is that what you were asking?

projectilemotion has a good formula for determining the distance with a single angle variable which is ... actually a better way to reduce the points (and their cooresponding distances from the center) to a single variable than to reduce it to a single $x$ or $y$ variable (which would fail to be a determinable function as each $x$ [except $0$] yields two $y$s and every $y$ [except $0$] yields two $x$s).