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.

somewhereon a plane(You know nothing about an invisible circle, so all you can do is to draw the first linesomewhere). Then,fortunatelyyou get a value $d_1$. Then, you know $l_1$ crossed an invisible circle at two points(This is all you can know.) Next, when you draw another line called $l_2$, againfortunatelyyou get a value $d_2$. (continued) $\endgroup$ – mathlove Aug 29 '13 at 9:417more comments