Finding an invisible circle by drawing another line A friend of mine taught me the following question. He said he found it in 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 equals $1$. 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\gt0)\ $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 give me how to find the coordinates?
The situation so far: The $l_1\parallel l_2$ case : This case has been already solved (see Blue's answer below).  
The $l_1\not \parallel l_2$ case : This case has not been solved yet. 
Supposing that   $l_1:y=x\tanθ$, $l_2:y=-x\tanθ$ and $l_3:y=0$ ($l_4:x=0$ if needed) for $0<θ<\pi/2$, then we can get two possible coordinates as the center of a circle. However, it seems difficult to decide just one coodinates because each line is symmetric about the origin.
Hence, a new line, which is not $y=0$, is needed as $l_3$.
My approach:
Let each of $l_{1,d+}, l_{1,d-}, l_{2,D+}, l_{2,D-}$ be the followings:$$l_{1,d+}:y=x\tanθ+\frac{d}{\cosθ}, l_{1,d-}:y=x\tanθ-\frac{d}{\cosθ}$$
$$l_{2,D+}:y=-x\tanθ+\frac{D}{\cosθ}, l_{2,D-}:y=-x\tanθ-\frac{D}{\cosθ},$$
where $D=\sqrt{d^2+\frac{{d_1}^2-{d_2}^2}{4}}.$
Note that each distance between $l_1$ and $l_{1,d\pm}$ is $d$, and that each distance between $l_2$ and $l_{2,D\pm}$ is $D$. Also, note the following:
$$\sqrt{\left(\frac{d_1}{2}\right)^2+d^2}=\sqrt{\left(\frac{d_2}{2}\right)^2+D^2}.$$
This means that the radius of a circle which crosses $l_1$ equals the radius of a circle which crosses $l_2$. Note that $d$ must satisfy the following:$$0\le d\le \sqrt{1-\frac{{d_1}^2}{4}}.$$
Then, Letting each of the intersections of $l_{1,d-}$ and $l_{2,D+}$, $l_{1,d+}$ and $l_{2,D+}$, $l_{1,d+}$ and $l_{2,D-}$, $l_{1,d-}$ and $l_{2,D-}$ be $P_{-+}$, $P_{++}$, $P_{+-}$, $P_{--}$ respectively, we can represent these as the follwoings:
$$P_{-+}\ \left(\frac{d+D}{2\sinθ}, \frac{-d+D}{2\cosθ}\right), P_{++}\ \left(\frac{-d+D}{2\sinθ}\frac{d+D}{2\cosθ}\right),$$$$P_{+-}\ \left(\frac{-d-D}{2\sinθ}, \frac{d-D}{2\cosθ}\right), P_{--}\ \left(\frac{d-D}{2\sinθ}, \frac{-d-D}{2\cosθ}\right).$$
Since each radius is $\sqrt{d^2+\frac{{d_1}^2}{4}}$, we can represent the circles by $d$ as the followings:
$$C_{-+}:\left(x-\frac{d+D}{2\sinθ}\right)^2+\left(y-\frac{-d+D}{2\cosθ}\right)^2=d^2+\frac{{d_1}^2}{4}$$
$$C_{++}:\left(x-\frac{-d+D}{2\sinθ}\right)^2+\left(y-\frac{d+D}{2\cosθ}\right)^2=d^2+\frac{{d_1}^2}{4}$$
$$C_{+-}:\left(x-\frac{-d-D}{2\sinθ}\right)^2+\left(y-\frac{d-D}{2\cosθ}\right)^2=d^2+\frac{{d_1}^2}{4}$$
$$C_{--}:\left(x-\frac{d-D}{2\sinθ}\right)^2+\left(y-\frac{-d-D}{2\cosθ}\right)^2=d^2+\frac{{d_1}^2}{4}.$$
Changing $d$ to $-d$ in $C_{++}$ gives $C_{-+}$ and changing $d$ to $-d$ in $C_{+-}$ gives $C_{--}$. Hence, we can represent each possible invisible circle by $d$ as the following:
$$C_{\pm+}:\left(x-\frac{-d+D}{2\sinθ}\right)^2+\left(y-\frac{d+D}{2\cosθ}\right)^2=d^2+\frac{{d_1}^2}{4}$$
$$C_{\pm-}:\left(x-\frac{-d-D}{2\sinθ}\right)^2+\left(y-\frac{d-D}{2\cosθ}\right)^2=d^2+\frac{{d_1}^2}{4}$$
for $d$ which satisfies the following:
$$-\sqrt{1-\frac{{d_1}^2}{4}}\le d\le \sqrt{1-\frac{{d_1}^2}{4}}.$$
In addition to this, letting $(x,y)$ be the center of each circle, we get the following:
$$xy=\frac{{d_1}^2-{d_2}^2}{16\cosθ\sinθ}.$$
This shows that the center of each possible invisible circle is on this hyperbola if $d_1-d_2>0$.
I've tried to get a special line as $l_3$, but I'm facing difficulty.
update: I crossposted to MO.
https://mathoverflow.net/questions/140435/finding-an-invisible-circle-by-drawing-another-line
 A: After struggling with equations for a few fruitless minutes, I went back and read the question more carefully ... ;-)
My solution is at https://mathoverflow.net/questions/140435/finding-an-invisible-circle-by-drawing-another-line/ where the problem was re-posted.
edit: In response to Daniel's request, the following is a copy & paste of my reply at mathoverflow (Good idea actually Daniel because it may be closed/deleted from there.)
begin quote
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. (Of course these circles will coincide if $d_1 = 2$.)
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!
end quote
P.S. I didn't elaborate the solution, because elementary geometric calculations seem out of place there, and I'd guess most people here would also have no difficulty in deriving the results explicitly based on this approach.
A: Trying another tack, let us denote the two given lines by $p_i x + q_i y = r_i$, where $p_i^2 + q_i^2 = 1$ ($i = 1, 2$) and, to avoid carting around a load of 2s, denote the respective chord lengths by $2 d_i$.
Then, denoting the centre of the circle by (u, v), which is to be found, the equation of the circle is as follows, in which $p u + q v = r$ is our 3rd (or 4th) line, again as always with p^2 + q^2 = 1 :
$(x - u)^2 + (y - v)^2 = (p_1 u + q_1 v - r_1)^2 + d_1^2 = (p_2 u + q_2 v - r_2)^2 + d_2^2 = (p u + q v - r)^2 + d^2$
Subtracting, to eliminate the leftmost sum of squares involving $x, y$, gives two conics (in general hyperbolas) in $u, v$. So if the question has a solution, then it must be possible to choose $p, q, r$ such that if d > 0 then these two conics intersect in exactly one point. This point must be at a common tangent if at least one conic is non-degenerate, or otherwise, i.e. if each conic is a pair of lines, one common intersection of all four lines.
The condition(s) on $p, q, r$ must not involve $d$, because that is not known in advance. But we can use different conditions based on the values of the $d_i$, for example whether or not $d_1 = d_2$.
I'll study this further tonight, but in the meanwhile feel free to edit/append the above if inspiration strikes!
