A friend of mine taught me the following question. He said he found it on a book a few years ago. Though I've tried to solve it, I'm facing difficulty.
Question: You know on a plane there is an invisible circle whose radius is less than or equal to $1$. Fortunately, you have already found that the lengths of the chords of a circle by two lines $l_1, l_2$ are $d_1, d_2$ ($2\gt d_1\ge d_2\gt 0$) respectively. By drawing another line, let's find this circle. If the line you'll draw crosses a circle at two points, then you'll get the length of the chord of a circle by the line. If the line you'll draw and a circle come in contact with each other, then you'll get the coordinates of the point of contact instead of getting $0$ as the length of the chord. If the line you'll draw neither crosses nor comes in contact with any circle, then you'll be able to draw another line just once more. Find the coordinates of the center of a circle.
This is all the question says. Could you show me how to find the coordinates?
This question has been asked previously on math.SE without receiving any complete answer:
https://math.stackexchange.com/questions/468324/finding-an-invisible-circle-by-drawing-another-line
The $l_1\parallel l_2$ case has been already solved (see Blue's answer on math.SE). On the other hand, the $l_1\not \parallel l_2$ case has not been solved yet.
I'm going to write several things about the $l_1\not \parallel l_2$ case which we've already found on math.SE. For further details, please see the page on math.SE.
In the following, suppose that $l_1:y=x\tanθ$, $l_2:y=-x\tanθ$ for $0<θ<\pi/2$ and that $a=\frac{d_1}{2}, b=\frac{d_2}{2}$.
1. Taking $l_3:y=0$ ($l_4:x=0$ if needed), then we can get two possible coordinates as the center of a circle. However, it seems difficult to decide just one coordinates because each line is symmetric about the origin. Hence, a new line, which is not $y=0$, is needed as $l_3$.
2. We can represent every possible invisible circle as the following:
$$C_{\pm+}:\left(x-\frac{-d+{\sqrt{d^2+a^2-b^2}}}{2\sinθ}\right)^2+\left(y-\frac{d+{\sqrt{d^2+a^2-b^2}}}{2\cosθ}\right)^2=d^2+a^2$$
$$C_{\pm-}:\left(x-\frac{-d-{\sqrt{d^2+a^2-b^2}}}{2\sinθ}\right)^2+\left(y-\frac{d-{\sqrt{d^2+a^2-b^2}}}{2\cosθ}\right)^2=d^2+a^2$$
where $d$ satisfies the following:
$$-\sqrt{1-a^2}\le d\le \sqrt{1-a^2}.$$
Hence, we know that the center of each circle is on the following hyperbola:
$$xy=\frac{a^2-b^2}{4\cosθ\sinθ}.$$
if $d_1-d_2>0$.
3. Blue got a quartic in $h$ with $\phi, p, c$ supposing that $l_3:x\sin\phi−y\cos\phi+p=0$ cuts a chord of length $2c$ in the circle with center $(h,k)$ and radius $r$.
4. In Blue's quartic, taking $\phi=0, \frac{\pi}{2}$ don't work in general because such lines don't necessarily hit every circle in a given sub-family of circles.
Update: I'm going to write my idea. I hope this would be helpful.
First, let's call the following circles 'upper-right sub-family of circles':
$$C_{\pm+}:\left(x-\frac{-d+{\sqrt{d^2+a^2-b^2}}}{2\sinθ}\right)^2+\left(y-\frac{d+{\sqrt{d^2+a^2-b^2}}}{2\cosθ}\right)^2=d^2+a^2.$$
Also, let's call the following circles 'lower-left sub-family of circles':
$$C_{\pm-}:\left(x-\frac{-d-{\sqrt{d^2+a^2-b^2}}}{2\sinθ}\right)^2+\left(y-\frac{d-{\sqrt{d^2+a^2-b^2}}}{2\cosθ}\right)^2=d^2+a^2$$
where $d$ satisfies $-\sqrt{1-a^2}\le d\le \sqrt{1-a^2}$.
We can see that each center of upper-right sub-family is in the first quadrant, and that each center of lower-left sub-family is in the third.
I've been looking for a special line $L$ which satisfies the following three conditions. If we can find such line, we can take the line as $l_3$.
1. $L$ crosses every circle of upper-right sub-family.
2. $L$ never crosses any circle of lower-left sub-family.
3. Each length of the chord of each circle of upper-right sub-family by $L$ is different from each other.
Now let $l_3$ be $x\sin\phi-y\cos\phi+p=0$. Actually, we can write these three as the following:
1. $|\frac{-d+{\sqrt{d^2+a^2-b^2}}}{2\sinθ}\sin\phi-\frac{d+{\sqrt{d^2+a^2-b^2}}}{2\cosθ}\cos\phi+p|\lt \sqrt{d^2+a^2}$ for any $d$.
2. $|\frac{-d-{\sqrt{d^2+a^2-b^2}}}{2\sinθ}\sin\phi-\frac{d-{\sqrt{d^2+a^2-b^2}}}{2\cosθ}\cos\phi+p|\gt \sqrt{d^2+a^2}$ for any $d$.
3. $f(d)=\sqrt{(d^2+a^2)-|\frac{-d+{\sqrt{d^2+a^2-b^2}}}{2\sinθ}\sin\phi-\frac{d+{\sqrt{d^2+a^2-b^2}}}{2\cosθ}\cos\phi+p|^2}$ is monotone increasing or decreasing function.
Then, we can say that if there exists a set of $(\phi,p)$ which satisfies these three conditions, then we can take $l_3$ to be the line $x\sin\phi-y\cos\phi+p=0$.
If this line doesn't cross any circle, then we can take $l_4$ to be the line which is origin-symmetric to $l_3$.
However, I neither know if there exists such $(\phi,p)$ for any $(\theta, a, b)$, nor know how to get such $(\phi, p)$ if it exists.
Best Answer
This I think is the big clue: " If the line you'll draw and a circle come in contact with each other, then you'll get the coordinates of the point of contact instead of getting 0 as the length of the chord."
If the first line cuts a chord of length $d_1$ then the circle is enclosed in a band centred on the line, ranging from being centred on the line when the circle has diameter $d_1$ to a unit circle offset on either side of the line, and the lines bounding the band are tangents to both these unit circles.
With that in mind, what you should do is draw the third line parallel to the first at a distance of $\frac{d_1}{2}$ from it. If the circle is offset the other side of the line, you then have a second chance, and you draw another parallel line the same distance the other side.
One of these lines will then be either tangent to the circle (if it is centred on the line and has diameter $d_1$), and in this case the coordinates of the point of contact easily allow one to deduce the circle's centre.
Otherwise one of these lines will cut out a chord, and the length of this combined with the original chord length and the distance apart of the parallel lines will allow the radius of the circle to be determined.
But once you know the circle's radius, the chord lengths cut by any two oblique lines allow its centre to be determined.
Very nice problem!