Intersecting waves of rain droplets Let us image the rainy weather and water droplets on the puddle. The waves that rain make finally intersect but let us define how do they intersect exactly. Let us say that we examine two water drops in particular. They are, mathematically speaking, points on the plane: the first second they are tiny drops on the puddle. But then they cause waves to expand, i.e. from those two points two distinct circles appear with increasing radius until they intersect.
Let us say we know that the 1st second, 1st drop was on (0,0) point and 2nd drop on (10,7). Then next second, we know that 1st drop transformed into circle with radius of $\sqrt2$ and 2nd drop transformed into circle with radius of $\sqrt3$. The 3rd second we know, that 1st circle is with radius of 2$\sqrt2$ and 2nd circle is with radius of 2$\sqrt3$. The question is:
After how many seconds the circles will come intact?
Maybe differential equation involving functions $R_1(t)$ and $R_2(t)$ would be applicable where $R_i$ is a radius of the $i$-th circle ($i=1,2$), $t$-time and we are given initial conditions of ODE.

Let's answer the question more generally: place the first drop at the origin with radius expanding at speed $u$ and the second drop at some $(a, b)$ in the plane, expanding at some speed $v$.
Try to fill in the blanks and click on any of the spoilers to reveal the answer.
At time $t$ measured after the drops splash simultaneously the radius of the first ring is

$$ut$$

and the radius of the second ring is

$$vt.$$

The Pythagorean theorem shows that the distance $r$ between the initial points satisfies

$$r^2 = a^2 + b^2,$$

hence at time $t$ the wave fronts moving towards one another are at a distance

$$r - ut - vt = r - (u+v)t$$.

When the wave fronts collide, this distance is $0$ hence we must solve the equation

$$r - (u+v)t = 0$$

for the variable $t$ which gives

$$t = \frac{r}{u+v} = \frac{\sqrt{a^2 + b^2}}{u+v}.$$


With your example values, $(a, b) = (10, 7)$ and $(u, v) = \bigl( \sqrt{2}, \sqrt{3}\, \bigr)$, hence the time until contact is

$$ t = \frac{\sqrt{10^2 + 7^2}}{\sqrt{2} + \sqrt{3}} \approx 3.88\ldots $$

