Related rates circle problem Two circles $A$ and $B$ have the same center. The radius of the inner circle $A$ is increasing at a rate of $1$ unit/sec, and the radius of the larger circle $B$ is also increasing such that the area between the two circles is always $10\pi$. When the radius of $A$ is 5, how fast is the radius of $B$ increasing?
I know I need to start by setting up an equation to perform related rates, but what that equation needs to be and how I need to solve it, I don't know (Note: I understand the topic, this is not that kind of issue).
 A: Let $R, r$ be the radii of the larger and smaller circles, respectively. Then:
$$\pi R^2-\pi r^2=10 \pi \Rightarrow R^2-r^2=10.$$
Differentiate with respect to time $t$:
$$2RR'-2rr'=0 \Rightarrow R'=\frac{rr'}{R}=\frac{5\cdot 1}{\sqrt{10+5^2}}=\frac{\sqrt{35}}{7}. $$
A: I always like to start these with a picture:

The area between A and B is just the area of B less the area of A. This is just $10\pi = \pi ((a+c)^{2}-a^{2})$ (Notice that the radius of B here is $a+c$).
We're free to divide out $\pi$ and simplify to get $10 = 2ac +c^{2}$. Differentiating both sides with respect to time we get $0 = 2a'c + 2ac' + 2cc'$. We can divide out 2. We already know that $a' = 1$. So we reduce to $0 = c + ac' + cc'$.
We are interested specifically when $a=5$, so we can substitute that in: $0=c+5c'+cc'$. To finish we need to compute what $c$ is when $a=5$. This isn't so hard. Just use the same equation as we started with and solve for $c$, $10 = (5+c)^{2} - 25$, $c = \sqrt{35}-5$. We put this value into our previous result and we're left with $0 = \sqrt{35}-5 + 5c' + (\sqrt{35}-5)c'$, so $c' = \frac{\sqrt{35}-5}{\sqrt{35}}$.
Remember that the rate of change of the radius of $B$ is actually $a' + c' = 1 +\frac{\sqrt{35}-5}{\sqrt{35}}$.
I hope there are no mistakes.
