Tag: circles

Choose any point outside a circle. Can we draw a second circle, centered at that point, that covers exactly half the area of the original circle?

Surprisingly, the answer is always yes. Even better, there is exactly one such circle.

Let the original circle have center O and radius R. Choose a fixed point P outside the circle, and let

$$d=OP>R.$$

We want to find the radius r of a circle centered at P that covers exactly half of the original disk.

Why must such a circle exist?

Before calculating anything, we can prove that the desired circle must exist.

Let A(r) denote the area of the part of the original disk covered by the circle centered at P with radius r.

Since P is outside the original circle, the distance from P to the nearest point of the original circle is

$$d-R.$$

Therefore, when

$$r=d-R,$$

the two circles are externally tangent. They meet at only one point, so their common area is zero:

$$A(d-R)=0.$$

Now continue increasing r. Eventually,

$$r=d+R.$$

At this point the circle centered at P contains the entire original disk. Hence

$$A(d+R)=\pi R^2.$$

As r increases, the circle centered at P grows continuously. Consequently, the area A(r) that it covers inside the original circle also changes continuously.

It starts at

$$0$$

and eventually reaches

$$\pi R^2.$$

Therefore, by the Intermediate Value Theorem, at some intermediate radius it must equal

$$\frac{1}{2}\pi R^2.$$

Thus there is at least one radius satisfying

$$d-R<r<d+R$$

for which the second circle covers exactly half of the original disk.

Why is the radius unique?

Suppose

$$r{}_1<r{}_2.$$

The disk centered at P with the smaller radius is strictly contained in the disk with the larger radius. While the two circles intersect, increasing the radius captures an additional region of positive area inside the original disk.

Therefore A(r) is strictly increasing for

$$d-R<r<d+R.$$

So the value

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

can occur only once. The required circle is therefore unique.

Now let us calculate its radius

Let the two circles intersect at points Q and S. Consider the triangle formed by O, P, and Q.

Its side lengths are

$$OQ=R,\hspace{1em}PQ=r,\hspace{1em}OP=d.$$

Let α be the angle at O between OP and OQ, and let β be the angle at P between PO and PQ.

By the Law of Cosines,

$$\cos \alpha =\frac{d^2+R^2-r^2}{2dR}.$$

Hence

$$\alpha =\arccos \left(\frac{d^2+R^2-r^2}{2dR}\right).$$

Applying the Law of Cosines at P gives

$$\cos \beta =\frac{d^2+r^2-R^2}{2dr},$$

so

$$\beta =\arccos \left(\frac{d^2+r^2-R^2}{2dr}\right).$$

The area of the overlap

The common lens consists of two circular segments.

At O, the full central angle subtending the common chord is 2α. The area of the corresponding sector is

$$R^2\alpha .$$

The triangle inside this sector has area

$$R^2\sin \alpha \cos \alpha .$$

Thus the first circular segment has area

$$R^2(\alpha -\sin \alpha \cos \alpha ).$$

Similarly, the second circular segment has area

$$r^2(\beta -\sin \beta \cos \beta ).$$

Therefore the total overlap area is

$$A\left(r\right)=R^2(\alpha -\sin \alpha \cos \alpha )+r^2(\beta -\sin \beta \cos \beta ).$$

The equation for the radius

We want the overlap to be exactly half the area of the original circle. Therefore r must satisfy

$$R^2(\alpha -\sin \alpha \cos \alpha )+r^2(\beta -\sin \beta \cos \beta )=\frac{1}{2}\pi R^2,$$

where

$$\alpha =\arccos \left(\frac{d^2+R^2-r^2}{2dR}\right)$$

and

$$\beta =\arccos \left(\frac{d^2+r^2-R^2}{2dr}\right).$$

This is a transcendental equation, so in general the radius is found numerically. But our geometric argument has already established something important: for every d > R, this equation has exactly one solution in

$$d-R<r<d+R.$$

A scale-free version

The problem really depends only on the ratio d/R. Define

$$D=\frac{d}{R},\hspace{1em}\rho =\frac{r}{R}.$$

Then D > 1, and the required value of ρ depends only on D. Once ρ is found, the actual radius is simply

$$r=\rho R.$$

For example, if the outside point happens to satisfy

$$d=2R,$$

the numerical solution is

$$r\approx 2.08225R.$$

But this is only one example. The construction works for every point outside the original circle.

The geometric conclusion

We have proved the following result:

Given any point outside a circle, there exists a unique circle centered at that point whose intersection with the original disk has exactly half the area of the original disk.

The calculation of its radius requires solving a transcendental equation, but its existence does not. It follows from a simple geometric idea: start with a circle that barely touches the original circle and continuously increase its radius. The covered area grows continuously from zero to the entire area of the original disk, and therefore it must pass through exactly one-half.

Some numerical examples

Because the problem is unchanged by scaling, we may take

$$R=1.$$

For several choices of the distance d from the center of the original circle to the outside point, the table below gives the unique radius r that covers exactly half of the original disk.

Distance d	Required radius r
1.10	1.24543
1.25	1.37910
1.50	1.60860
2.00	2.08225
2.50	2.56611
3.00	3.05523
5.00	5.03326

For a circle of arbitrary radius R, these numbers should be interpreted as ratios. For example,

$$\frac{d}{R}=2$$

gives

$$\frac{r}{R}\approx 2.08225,$$

or equivalently,

$$r\approx 2.08225R.$$

The table also reveals an interesting pattern: as the outside point moves farther from the original circle, the required radius becomes increasingly close to the distance d itself.

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Another Equal-Area Problem

Dividing the area of a circle exactly in half with another circle raises a natural geometric question: can a circle do the same thing to a triangle?

Continue exploring: Can a Circle Cut a Triangle in Half?