Area between two circles equal to half the area of one of the circles

Question

Consider the above picture, where you would want to find a value for $r$, depending on $R$ so that the conjoint area (marked gray) is equal to half the area of the $\color{darkorange}{\text{orange circle}}$, i.e.

$$A(r) = \frac{1}{2}\pi R^2$$

Come up with a way to find $r$, if the $\color{blue}{\text{blue circle's}}$ centre is located on the perimeter of the $\color{darkorange}{\text{orange circle}}$.

---

I'm okay with using Mathematica if necessary, I just can't seem to find a way to solve my problem. Also allow me to streamline the wording of the problem.

Problem worded differently

If you have a circle with radius $R$, and you attach another circle, with its centre, on the perimeter of the first circle. Find the radius $r$ (see first image) for the second circle such that the area between the circles is $50\%$ of the area of the first circle.

Answer 1

I don't use analytic geometry.

Let $A$ and $B$ be the points of intersection of the circles and $C$ the center of the circle of radius $r$.

Let $x=\angle ACB$ our unknown. Then $$r=2R \cos \frac x2$$ Simple facts about geometry and trigonometry give the equation $$\sin x - x \cos x= \frac \pi2$$ so $x$ is approximately $1.9056957$ and $r/R$ is approximately $1.1587285$ .

Answer 2

You can do it without calculus. Draw the line AB. The lens is composed of two circular segments, one generated by each circle. If the center of the small circle is $C$ and the center of $AB$ is $D$, you have $AC=r, AD=\sqrt{R^2-(R-CD)^2}=\sqrt{r^2-CD^2}. $ Once you find the coordinates of $A$, you can just integrate $\sqrt{r^2-x^2}-(R-\sqrt{R^2-x^2})$ from $A$ to $B$ and have the area. This is much harder than using geometry, but you can get there.

Answer 3

Let $C_1$ be our $\color{blue}{\text{blue circle}}$, and $C_2$ our $\color{darkorange}{\text{orange circle}}$ with the following coordinates. Note that $C_2$ lies at $(0,R)$ which is an important detail for this problem.

Our two circles' equations are

$\begin{eqnarray} C_1&:& x^2+y^2 &= R^2 \tag{1}\\ C_2&:& (x-R)^2+y^2 &= r^2 \tag{2} \end{eqnarray}$

Combining $(1)$ and $(2)$ and solving for $x$ we get

$$(x-R)^2+(R^2-x^2) = r^2 \\ x_1 = \dfrac{2R^2-r^2}{2R}$$

So what did we just find? We found the $x$-coordinate at which the two circles intersect. Using the formula to find an area for a circular segment. Let $R'$ be a radius, and $d$ a distance from a circle's centre to a chord. For our case, our chord would be a vertical line going through the intersection points of our circles.

$$A(R',d) = R'^2\cdot\cos^{-1}\left(\frac{d}{R'}\right) - d\sqrt{R'^2-d^2}$$

Using this, and plugging in respectively value for our circles we get the following

$$A(R,x_1) = R^2\cdot\cos^{-1}\left(\frac{x_1}{R}\right) - x_1\sqrt{R^2-x_1^2}$$

So where are we now? We've calculated the shaded area below, and we need to find $d$ so that we can do the same for the other circular segment.

Well, we know that the distance between our circles is $R$. So why not put $d = R-x_1$. Hence,

$$A(r,d) = r^2\cdot\cos^{-1}\left(\frac{d}{r}\right) - d\sqrt{r^2-d^2}$$

This would mean our $A_{total} = A(R,x_1) + A(r,d)$. Writing this out becomes tedious, but using Mathematica gives,

$$A_{total} = r^2\cos^{-1}\left(\frac{r^2}{2R\cdot r}\right) + R^2\cos^{-1}\left(\frac{2R^2-r^2}{2R^2}\right) - \frac{1}{2}\sqrt{r^2\cdot 2R(2R+r)}$$

Now, using our $A_{total}$ we can find values for $r$ by specifying $R$ and the area ratio you want. I.e.

If $R = 2$ and you want the area between the circles to equal half of $C_1$'s area, you would put:

$$R = 2: A_{total} = \frac{1}{2}\pi\cdot R^2 = 2\pi$$

Let Mathematica calculate $A_{total}$ numerically after specifying $R$, and you will find it gives you two solutions. Use the positive numerical solution and that's the radius you were looking for.

NB! By trying different values for $R$ and dividing the result $r$ with $R$, respectively, you'll find $\frac{R}{r} \approx 1.15872847 \ldots$. (OEIS A133731)

And as such, for this particular problem, the radius $r$ is always near $1.16\cdot R$ length units.