Let's look at the number of the representations of $n$ as $a^2 + b^2$ with coprime $a, b$, and denote it $v(n)$. Our sum is then $\sum_{n \leq r^2} {\frac{v(n)}{n}}$. We can see that $\sum_{d^2 | n} v(\frac n{d^2}) = r_2(n)$, so $v(n) = \sum_{d^2 | n} r_2(\frac n{d^2}) \mu(d)$, therefore $$ \sum_{n \leq r^2} {\frac{v(n)}{n}} = \sum_{q d^2 \leq r^2} {\frac{r_2(q) \mu(d)}{q d^2}} = \sum_{q \leq r^2} \frac{r_2(q)}{q} \sum_{d \leq \frac{r}{\sqrt q}} \frac{\mu(d)}{d^2} = \sum_{q \leq r^2} \frac{r_2(q)}{q} (\frac{6}{\pi^2} + O(\frac{\sqrt{q}}{r})) = \frac{6}{\pi^2} \sum_{q \leq r^2} \frac{r_2(q)}{q} + O(1) = \frac{6}{\pi^2} \sum_{a^2 + b^2 \leq r^2} \frac{1}{a^2 + b^2} + O(1)$$
which shows it's $\frac{6}{\pi^2}$-th of the unrestricted sum plus $O(1)$, which following @Terry Tao's comment can be shown to be $2 \pi \log(r) + O(1)$, so the result is $\frac{12}{\pi} \log(r) + O(1)$.
I found a much simpler proof, below is the original one:
We'll use $r$ as the radius squared plus one, instead of the radius. Additionally, we'll only deal with the first quadrant, and at the end we can multiply the result by 4. $$\sum_a {\sum_{b\\(a,b)=1\\a^2 + b^2 < r}} \frac{1}{a^2 + b^2} = 1 + 2\sum_a {\sum_{b < a, \sqrt{r - a^2} \\(a,b)=1}} \frac{1}{a^2 + b^2} = 1 + 2(\sum_{a \leq \sqrt{\frac r2}} {\sum_{b < a \\(a,b)=1}} \frac{1}{a^2 + b^2} + \sum_{\sqrt{\frac r2} < a < \sqrt r} {\sum_{b < \sqrt{r - a^2} \\(a,b)=1} \frac{1}{a^2 + b^2}})$$ we can notice that $$\sum_{b < \sqrt{r - a^2} \\(a,b)=1} \frac{1}{a^2 + b^2} < \sum_{b < \sqrt{r}} \frac{1}{a^2} \leq \frac{\sqrt{r}}{a^2}$$ so $$\sum_{\sqrt{\frac r2} < a < \sqrt r} {\sum_{b < \sqrt{r - a^2} \\(a,b)=1} \frac{1}{a^2 + b^2}} \leq \sqrt r \sum_{\sqrt{\frac r2} < a < \sqrt r} {\frac{1}{a^2}} \leq \sqrt{r} \sum_{\sqrt{\frac r2} < a < \sqrt r} {\frac2r} \leq 2$$
So our sum is $$O(1) + 2\sum_{a \leq \sqrt{\frac r2}} {\sum_{b < a \\(a,b)=1}} \frac{1}{a^2 + b^2}$$ Now we'll estimate ${\sum_{b < a \\(a,b)=1}} \frac{1}{a^2 + b^2}$: $$\sum_{b < a \\(a,b)=1} \frac{1}{a^2 + b^2} = \sum_{b \leq a} \sum_{q | a\\q | b}\mu(q)\frac{1}{a^2 + b^2} = \sum_{q | a} \mu(q) \sum_{b \leq \frac aq} \frac{1}{a^2 + (qb)^2} = \sum_{q | a} \frac{\mu(q)}{q^2} \sum_{b \leq \frac aq} \frac{1}{(\frac aq)^2 + b^2} = \sum_{q | a} \frac{\mu(q)}{q^2} (\int_0^{\frac aq} \frac{1}{(\frac aq)^2 + x^2} dx + O((\frac qa)^2)) = \frac{\pi}{4} \sum_{q | a} \frac{\mu(q)}{a q} + O(\frac {d(a)}{a^2}) = \frac{\pi}{4} \sum_{q | a} \frac{\mu(q)}{a q} + O(a^{-\frac32})$$ And therefore $$\sum_{a \leq \sqrt{\frac r2}} {\sum_{b < a \\(a,b)=1}} \frac{1}{a^2 + b^2} = \frac{\pi}{4} \sum_{qd \leq \sqrt{\frac r2}} \frac{\mu(q)}{q^2 d} + O(1) = \frac{\pi}{4} \sum_{q \leq \sqrt{\frac r2}} \frac{\mu(q)}{q^2} (\log(\frac{\sqrt{\frac r2}}{q}) + O(1)) + O(1) = \frac{\pi}{4} \left(\log(\sqrt{\frac r2}) \sum_{q \leq \sqrt{\frac r2}} \frac{\mu(q)}{q^2} - \sum_{q \leq \sqrt{\frac r2}} \frac{\mu(q) \log(q)}{q^2}\right) + O(1) = \frac{\pi}{4} \log(\sqrt{\frac r2}) (\frac{6}{\pi^2} + O(\sqrt\frac2r)) + O(1) = \frac{3}{2\pi} \log(\sqrt{r}) + O(1)$$ And the total illumination in the first quadrant is $\frac{3}{\pi} \log(\sqrt r) + O(1)$, and the total illumination in the entire circle is $\frac{12}{\pi} \log(\sqrt r) + O(1)$. Since we took $r$ as the radius squared, the answer in the original question's terms is $\frac{12}{\pi} \log(r) + O(1)$.
