# Circle inside circle

Here's a fun little math problem I made(by fun I mean no complicated methods needed at all.)

You have a large circle with radius N with a cross in it to seperate the 4 quadrants(so that the figure looks like an 'xor' symbol). You also have a small circle with radius N/5.

Question:

What is the probability that the small circle will intersect the cross(anywhere) when randomly placed inside the large circle?

Note: Even though this is indeed a math problem, I think the approach to the solution(at least the one I came up with) is quite nice. So give it a shot!

• Just to make sure, when the small circle is placed inside the large circle, this means that it is fully contained, not just that the center is within the circle? – Michael Moschella Jun 25 '20 at 12:16
• @MichaelMoschella, Yes, fully contained inside the large circle. – Prim3numbah Jun 25 '20 at 12:20
• Is the placement random uniform? – msh210 Jun 25 '20 at 14:00
• Even with an uniform distribution this is still an ill-defined problem. To figure out the location of the small circle's centre point, we can uniformly randomise (for example) 1) the direction and distance from the big circle's centre, or 2) the x and y coordinate (rerolling any pairs that miss the intended area), or 3) the endpoints of a chord whose midpoint we'd use for the small circle's centre. All these are valid ways to uniformly distribute a point into a circle, and all of them will give wildly different answers to this problem. – Bass Jun 25 '20 at 15:14
• @Bass Your first and third example aren't "uniform" distributions, because the density reduces as you approach the edges (unless you use r = R*sqrt( rand() ) for the first example, in which case it gives the same distribution as the second example does) – Chronocidal Jun 26 '20 at 9:35

There are two simplifications we should make first:

We wish to reduce the problem of placing a circle to that of placing a point. Since the small circle had raids $$\frac{N}{5}$$, the center lies in a circle of radius $$\frac{4N}{5}$$, and the set of centers corresponding to circles crossing the cross is the set of points within $$\frac{N}{5}$$ of it.

Then, it's a matter of calculation.

We work in an octant of the circle, of which one side belongs to the cross.
We draw a line $$L$$ parallel to that side at distance $$\frac{N}{5}$$, call the point where it intersects the circle $$I$$, and draw the radius $$OI$$.
$$OI$$ makes an angle of $$\theta = \arcsin(\frac{1}{4})$$ with the arm of the cross, so the sector defined by it has area $$8\theta\left(\frac{N}{5}\right)^2$$.
The remaining area, bounded by $$L$$, $$OI$$, and the radius of the octant off the cross forms half of a parallelogram with height $$\frac{N}{5}$$ and base $$\frac{(\sqrt{15}-1)N}{5}$$; its area is therefore $$\frac{\sqrt{15}-1}{2}\left(\frac{N}{5}\right)^2$$.
Adding eight copies of each and dividing by $$16\pi\left(\frac{N}{5}\right)^2$$ (the area of all possible centers) gives $$\frac{\sqrt{15}-1+16\arcsin(\frac{1}{4})}{4\pi}$$ for our probability, or approximately 55%.

• Some graphics would be nice. Somehow, however, I miss the "quite nice approach" of the original author. Makes me wonder... – BmyGuest Jun 25 '20 at 13:13
• Added a figure. Is that how you meant? Aren't you double-counting the central overlapping squares (N/5)*(N/5) in your answer? – BmyGuest Jun 25 '20 at 13:33
• There is a square of width 2N/5 at the center of the circle that seems to be double-counted (each quarter of the square is part of two of your octant calculations). Did I miss where you took that into account? – user3294068 Jun 25 '20 at 16:55
• @user3294068 The image edited in by BmyGuest showed a purple Parallelogram which was not the same one that AxiomaticSystem was calculating, and showed that overlap. The image has been updated with the correct Parallelogram, so it should now be clear that there is no double-counting. (Also, it was a Rhombus, not a Square) – Chronocidal Jun 25 '20 at 18:02
• @Chronocidal Thx for the edit. – BmyGuest Jun 25 '20 at 18:48

here is my diagram shows how it should be solved easily;

This question becomes simply a geometry question which requires to calculate the area ratio between

the area $$\widetilde{FGH}$$ and and $$\widetilde{IAJ}$$

where it shows you can put your small circles in this big circular area.

the area of $$\widetilde{IAJ}$$

$$\widetilde{IAJ}$$ is simply one in four of $$\pi \left (\frac{4N}{5}\right )^2$$

and the area of $$\widetilde{FGH}$$ is a little more serious and calculated by @AxiomaticSystem resulted the ratio as

around 55%

Note that @AxiomaticSystem already solved this right, I just simplified what he has done to be understood easily.

Since the small circle is to be contained in the large one, its center must be at a random point inside the circle of radius 4N/5, and must be at least a distance of N/5 away from the cross. This means that the area in which the circle will not intersect the cross corresponds to 4 quarter circles of radius 3N/5, or just a full circle of the same radius, therefore the probability of not intersecting the cross would be equal to $$pi(3N/5)^2/pi(4N/5)^2$$, or simply 9/16. Inverting this we get the odds of intersecting the cross to be 7/16.

• I was thinking the same at first, but the curvature of the small circle would not match... Or am I missing something? see figure – BmyGuest Jun 25 '20 at 12:25
• Ah I see my problem. I incorrectly judged that at the nearest point not intersecting the cross, it would be N/5 from the center, when it's in reality (sqrt(2)/2)(N/5) since it's N/5 from each edge – Michael Moschella Jun 25 '20 at 12:34
• Taking away the bands of width N/5 doesn't give you a quarter-circle, either – AxiomaticSystem Jun 25 '20 at 12:46

Answer: $$\frac{16}{25}$$.

Let's suppose that cross coincides with the axes x/y in the two-dimensional plane. Consider positive quadrant. You can easily see that if the center of the $$\frac{1}{5}R$$ circle falls into an inscribed part of smaller circle of radius $$\frac{3}{5}R$$ described as : $$\{ x^2 + y^2 \le (\frac{4}{5}R)^2; x>\frac{R}{5}, y>\frac{R}{5}\}$$, then the circle will not intersect the axes., I.e. probability there is no intersection is $$(\frac{3}{5}R)^2/R^2 = \frac{9}{25}$$. Hence, probability to intersect is $$\frac{16}{25}$$..

• The area in which the centre can fall must necessarily have the same curvature on the arc as the full-sized circle, not a tighter curvature, which means it is not a quarter-circle. See Oray's diagram, or the first comment on Michael Moschella's answer for visual clarification – Chronocidal Jun 29 '20 at 8:17