So far I approached this problem computationally, I decided to take all the radii add them up, by distance formula, then divide by the number of radaii. To make the distribution even, I rotated the lines for a specific angle ranging from $0<h<2\pi$
I used the equation $\frac{x^2}{9}+\frac{y^2}{4}=1$
I realized I could do this using summations, and divide it by all the number radii.
All my work can be shown at https://www.desmos.com/calculator/8oynrehlep
There is probably a much easier way? Can anyone show me?
This is a little more advanced than the tags you put on this question, but in pre-calculus you'll learn the sinusoidal functions and polar graphing, where the dependant coordinate is exactly what you're after; radius. It's possible to convert a rectangular function to polar, using the following identities:
$$x = r*\cos{\theta}$$ $$y=r*\sin{\theta}$$ $$r^2=x^2 +y^2$$
That last one is directly from the distance formula you've been using, in turn based on the Pythagorean theorem (for any right triangle with sides of length $a,b$ and hypotenuse $c$, $c^2 = a^2 + b^2$).
So, let's plug those in to your equation and solve for r:
$$\dfrac{(r\cos\theta)^2}{9} + \dfrac{(r\sin\theta)^2}{4} = 1$$ $$\dfrac{r^2\cos^2\theta}{9} + \dfrac{r^2\sin^2\theta}{4} = 1$$ $$r^2(\dfrac{\cos^2\theta}{9} + \dfrac{\sin^2\theta}{4}) = 1$$ $$\dfrac{r^2}{36}(4\cos^2\theta + 9\sin^2\theta) = 1$$ $$\dfrac{r^2}{36} = \dfrac{1}{4\cos^2\theta + 9\sin^2\theta}$$ $$r^2 = \dfrac{36}{4\cos^2\theta + 9\sin^2\theta}$$ $$r = \sqrt{\dfrac{36}{4\cos^2\theta + 9\sin^2\theta}}$$ $$r = \dfrac{6}{\sqrt{4\cos^2\theta + 9\sin^2\theta}}$$
Now, you can plug in incrementally increasing angle values from 0 to 2$\pi$ radians (or set your calculator to degrees and go from 0 to 360) and average those values.
Now, if you want to get into even higher math, there is a defined formula for the average value of a function over a given domain. It is related to the definite integral of a function over the same domain, and for $r=f(\theta)$ over the angle domain $\alpha\leq\theta\leq\beta$ it looks like:
$$r_{av} = \dfrac{1}{\beta-\alpha}\int_\alpha^\beta f(\theta)d\theta$$
So, calculate the following definite integral and you'll have an exact answer:
$$\dfrac{1}{2\pi}\int_0^{2\pi} \dfrac{6}{\sqrt{4\cos^2\theta + 9\sin^2\theta}} d\theta$$
That's not an easy one, to be sure; at least three levels of compounding requiring nested integration by substitution. I didn't get that right (been 15 years since I've had that kind of chops in integral calculus) and ended up with an indefinite integral containing a $\sin\theta\cos\theta$ term in the denominator, making the integral undefined at any $\theta=2k\pi$. So, I just cheated and plugged it into a graphing calculator, and got 15.2339 for the definite integral, divided by $2\pi$ to give an average radius of 2.4246, which is a sensical value given the ellipse in question.