Finding an invisible circle by drawing another line

Solution 1:

To avoid some subscripts and fractions, I define $a := \frac{1}{2}d_1$ and $b := \frac{1}{2}d_2$. Also, I'll take the (unknown) center of the circle to be $(h, k)$, and its (unknown) radius to be $r$.

The $\ell_1 \parallel \ell_2$ Case. (Solved!)

With an appropriate change of variables, we can take $\ell_1$ to be $y = t$, and $\ell_2$ to be $y =-t$, for some $t > 0$.

The distance from $(h,k)$ to $\ell_1$ (or $\ell_2$) is $|k-t|$ (respectively, $|k+t|$), and Pythagoras tells us $$a^2 + |k-t|^2 = r^2 = b^2 + |k+t|^2$$ We can solve the "outer" equality to get $$k = \frac{1}{4t}\left( a^2 - b^2 \right)$$ so that $$r = \frac{1}{4t}\sqrt{\left( \left( a - b \right)^2 + 4 t^2 \right) \left( \left( a + b \right)^2 + 4 t^2 \right)}$$

Now knowing $k$ and $r$, we're guaranteed that the line $y=k+r$ will be tangent to the circle. Taking this as our $\ell_3$, the puzzle reports the point-of-tangency's coordinates, the first of which is our sought-after $h$. The center of the circle has been found!

The $\ell_1 \not\parallel \ell_2$ Case. (Incomplete.) Edited to remove false starts.

Here, we can take $\ell_1$ to be $y =x\tan\theta$, and $\ell_2$ to be $y=-x \tan\theta$, for some $0 < \theta < \pi/2$.

The distance from $(h,k)$ to $\ell_1$ (or $\ell_2$) is $|h \sin\theta - k\cos\theta|$ (respectively, $|h\sin\theta+k\cos\theta|$), so that $$\begin{align} a^2 &= r^2 - \left( h \sin\theta - k \cos\theta \right)^2 \qquad (1)\\ b^2 &= r^2 - \left( h \sin\theta + k \cos\theta \right)^2 \qquad (2) \end{align}$$

Note that subtracting $(2)$ from $(1)$ gives $$h k =\frac{a^2 - b^2}{4\cos\theta\sin\theta} \qquad (\star)$$ so that the centers of potential solution circles lie on a rectangular hyperbola.

Now, $(1)$ and $(2)$ are but two equations in three unknowns $(h, k, r)$, so that we have a one-parameter family of possible solutions. Here's a diagram of a typical family (with two clear sub-families we'll call $L$(eft) and $R$(ight)):

The four largest circles have radius $1$; the two smallest circles have radius $a$. The circles with radius $r$ have centers satisfying $(\star)$: $$(h, k) = \left( \frac{\pm_1\sqrt{r^2-b^2}\pm_2\sqrt{r^2-a^2}}{2\sin\theta}, \frac{\pm_1\sqrt{r^2-b^2}\mp_2\sqrt{r^2-a^2}}{2\cos\theta} \right)$$

There's no common tangent to help us, so the challenge would appear to be to take $\ell_3$ a line that cuts through $R$ (missing $L$) in such a way that its intersection with each circle creates a distinct chord-length. (And $\ell_4$ would be the corresponding line through $L$.)

So, let's try.

Let $\ell_3$ be the line with equation $x \sin\phi - y \cos\phi + p = 0$. The distance from $(h,k)$ to the line is $|h \sin\phi - k \cos\phi + p|$, and if the line cuts a chord of length $2c$ in the circle with center $(h,k)$ and radius $r$, then $$c^2 = r^2 - ( h \sin\phi - k \cos\phi + p )^2 \qquad (3)$$

Eliminating $r$ and $k$ from $(1)$, $(2)$, $(3)$ gives this quartic in $h$: $$\begin{align} 0 &= 16 h^4 \sin^2\theta \cos^2\theta \left( \sin^2\phi - \sin^2\theta \right) + 32 h^3 p \sin^2\theta \cos^2\theta \sin\phi \\ &- 8 h^2 \sin\theta \cos\theta \left((a^2-b^2) \sin\phi \cos\phi + (a^2+b^2-2c^2-2p^2) \sin\theta \cos\theta \right) \\ &- 8 h p \sin\theta \cos\theta \cos\phi \left( a^2 - b^2 \right) - \left( a^2 - b^2 \right)^2 \left(\sin^2\phi - \sin^2\theta \right) \qquad (4) \end{align}$$

From here, the challenge is to choose $\phi$ and $p$ so that, for every semi-chord length $c$ in the interval $I := [0,1]$ (or possibly a sub-interval thereof), there is a unique root $h$ in the interval $$H := \left[\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2\sin\theta}, \frac{\sqrt{1-b^2}+\sqrt{1-a^2}}{2\sin\theta}\right]$$ and, moreover, every $h\in H$ is that unique root for some $c \in I$ (or the sub-interval thereof).

After considerable effort, I have not been able to meet this challenge. Here are some notes dismissing some convenient cases:

• Neither a vertical line nor a horizontal line leads to a "universal" solution. The width of interval $H$ is $\sqrt{1-a^2}/\sin\theta$. For small enough $\theta$, this width exceeds $2$, so that no vertical line simultaneously hits the left-most and right-most unit circles in $R$. Likewise, large enough $\theta$ thwarts any horizontal line.

• Switching from vertical to horizontal (or vice-versa) after a certain threshold won't work, either. For small enough $\theta$, the $L$'s left-most (unit) circle has its center very-nearly on the $x$-axis. For certain values of $a$ and $b$, a horizontal $\ell_3$ missing that circle will also miss smaller members of $R$.

• $\phi = \theta$ isn't a "universal" solution. For some values of $(a,b,\theta)$, lines through through $R$ hit $L$; for other values, as in the horizontal/vertical cases, the circles of a family can get separated enough to avoid simultaneous contact with any line with the given slope.

• $\phi = -\theta$ doesn't work Here, $\ell_3$ runs parallel to $\ell_2$, which we know hits the unit-circle members (in fact, all members) of $R$ in identical-length chords; thus, $\ell_3$ must also meet those circles in identical-length chords, preventing chord length from determining a unique circle.

Additional thoughts on the non-parallel case.

As mentioned, every $h\in H$ must correspond to some $c$, and must be a root of $(4)$ with that $c$-value. Let the maximum and minimum endpoints of $H$ be $h_0$ and $h_1$, and let the corresponding $c$s be $c_0$ and $c_1$. (Observe that $(c_0,c_1) = (0,1)$ or $(1,0)$ represents the cases in which line $\ell_3$ is tangent to one of the unit circles in $R$ while passing through the center of the other.)

If there is a "universal" choice for $\phi$ and $p$ that works for particular parameters $(a,b,\theta)$ ---that is, if there are no "thresholds" at which we'd switch from one choice to another--- then by substituting $h=h_i$ and $c=c_i$ into $(4)$, we get two equations that we should be able to solve for $\phi$ and $p$.

So, let's try that.

I'll start by using these substitutions $$a^2 - b^2 = 4 \sin^2\theta h_0 h_1 \qquad a^2 + b^2 = 2\left( 1 - (h_0^2+h_1^2)\sin^2\theta \right)$$ to write $(4)$ thusly $$\begin{align} 0 &= h^4 \cos^2\theta \left( \sin^2\phi - \sin^2\theta \right) +2 h^3 p \cos^2\theta \sin\phi \\ &-h^2 \cos\theta \left( 2 h_0 h_1 \sin\theta \sin\phi \cos\phi + \cos\theta \left( 1 - c^2 - p^2 - (h_0^2+h_1^2) \sin^2\theta \right) \right) \\ &-2 h h_0 h_1 p \sin\theta \cos\theta \cos\phi - h_0^2 h_1^2 \sin^2\theta \left( \sin^2\phi - \sin^2\theta \right) \qquad (4^\prime) \end{align}$$ Also, I'll define first-quadrant angles $\gamma_0$ and $\gamma_1$ via $$\sin\gamma_0 = c_0 \qquad \sin\gamma_1 = c_1$$ Then, substituting $h=h_0$, $c=\sin\gamma_0$ and $h=h_1$, $c=\sin\gamma_1$ yields $$\begin{align} h_0 \cos\theta \sin\phi - h_1 \sin\theta \cos\phi + p \cos\theta &= \pm \cos\theta \sin\gamma_0 &(5a)\\ h_0 \sin\theta \cos\phi - h_1 \cos\theta \sin\phi - p \cos\theta &= \pm \cos\theta \sin\gamma_1 &(5b) \end{align}$$ Adding $(5b)$ from $(5a)$ gives this equation for $\phi$: $$\sin(\phi+\theta) = \frac{\pm_1\;\cos\theta}{h_1-h_0}\left(\sin\gamma_0\;\pm_2 \;\sin\gamma_1\right) = \frac{\pm_1\;\sin 2\theta}{\sqrt{1-a^2}}\sin\frac{\gamma_0 \pm_2 \gamma_1}{2}\cos\frac{\gamma_0 \mp_2 \gamma_1}{2} \qquad (6)$$ The two sign choices lead to four candidate values of $\phi$, which can be back-subsituted into the above equations to find $p$. Bear in mind that $\gamma_0$ and $\gamma_1$ (correspondingly, $c_0$ and $c_1$) aren't specified; adjusting their values undoubtedly affects the viability of the solution.

Edit. Let's take this a little further.

To avoid rampant "$\pm$"s in the formulas, define

$$\delta_0 = \pm_0\;\gamma_0 \qquad \delta_1 = \pm_1\;\gamma_1$$

Now, we can solve $(5a)$ and $(5b)$ as a linear system in $\sin\phi$ and $\cos\phi$, getting $$\begin{align} \cos\phi &= \frac{\cos\theta}{\sin\theta}\frac{(h_1-h_0)p - h_0\sin\delta_1 - h_1\sin\delta_0}{h_1^2-h_0^2} \\[6pt] \sin\phi &= -\frac{(h_1-h_0)p + h_0\sin\delta_0 + h_1\sin\delta_1}{h_1^2-h_0^2} \end{align}$$ The relation $\cos^2\phi + \sin^2\phi = 1$ then gives us this quadratic equation in $q := p ( h_1 - h_0 )$: $$\begin{align} 0 = q^2 &+ 2 q \left( \sin^2\theta ( h_0 \sin\delta_0 + h_1 \sin\delta_1 ) - \cos^2\theta(h_0\sin\delta_1+h_1\sin\delta_0 )\right) \\ &+ \sin^2\theta \left( h_0 \sin\delta_0 + h_1 \sin\delta_1 \right)^2 + \cos^2\theta \left( h_0 \sin\delta_1 + h_1\sin\delta_0 \right)^2 - \sin^2\theta \left(h_1-h_0\right)^2 \end{align}$$ with this discriminant $$\Delta := 4\left(h_0 + h_1 \right)^2 \sin^2\theta \left( (h_1-h_0)^2 - \cos^2\theta (\sin\delta_0 + \sin\delta_1 )\right)$$

Introducing first quadrant angles $\alpha$ and $\beta$ such that $$\sin\alpha = a \qquad \sin\beta = b$$ whence $$h_0 = \frac{\cos\beta - \cos\alpha}{2\sin\theta} \qquad h_1 = \frac{\cos\beta + \cos\alpha}{2\sin\theta}$$ we can ultimately write $$\begin{align} 2 p \cos\alpha &= \cos\alpha (\sin\delta_0 - \sin\delta_1 ) + \cos 2\theta\cos\beta(\sin\delta_0 + \sin\delta_1 ) \\[6pt] &\pm \cos\beta\sqrt{4\cos^2\alpha-\sin^2 2\theta (\sin\delta_0 + \sin\delta_1)^2 } \end{align}$$

Because the discriminant must be non-negative, we have this condition on $\delta_0$ and $\delta_1$: $$|\sin\delta_0 + \sin\delta_1| \leq \frac{2\cos\alpha}{\sin 2\theta} \qquad (\star\star)$$ This turns out to be equivalent to the requirement in $(6)$ that $|\sin(\phi+\theta)| \leq 1 $. Consequently, if we were to choose $\delta_0$ and $\delta_1$ in such a way as to make the discriminant vanish (which may or may not be possible), then we would have $\sin^2(\phi+\theta) = 1$, whereupon $\phi = \pi/2 - \theta$. In the case $\theta = \pi/4$, then, our line $\ell_3$ would coincide with $\ell_1$, giving no new information about our target circle. I take this to suggest that we shouldn't allow the discriminant to vanish in general.

Observe that $(\star\star)$ rules-out being able to choose $c_i \in \{0,1\}$ in some cases. So, the "take $\ell$ tangent to one unit circle and containing the center of the other" strategy won't always work.

Warning. Double-check all equations. I previously messed-up some signs; moreover, the LaTeX editing on this page has gotten so jerky that typos are very likely.

Solution 2:

Here is my answer. My answer is something like 'algorism' solution. Actually I'm not satisfied with my answer. Please improve my answer.

I think we can find a center even in the case $l_1\parallel l_2$.

Note that im my answer I suppose that we've already known the information about $l_1, l_2$.

In the case of $l_1\parallel l_2$, letting $d_0$ be the distance between two lines $l_1, l_2$, note that $0<d_0<2$ because of $0<d_2\le d_1<2$. Without loss of generality, suppose that $l_1:y=0, l_2:y=d_0$. Let $r$ be the radius of a circle. According to $0<d_2\le d_1<2$, there are two cases:case1 is that there exists the center of a circle between two lines and case2 is that there is not the center of a circle between two lines. In the first case, since we get $d_0=\sqrt{r^2-\left(\frac{d_2}{2}\right)^2}+\sqrt{r^2-\left(\frac{d_1}{2}\right)^2}$, we can represent $r$ by $d_0, d_1, d_2$. Then, letting $l_3:y=r+\sqrt{r^2-\left(\frac{d_1}{2}\right)^2}$, there are two small cases: one is that $l_3$ and a circle come in contact with each other and the other is that $l_3$ neither crosses nor come in contact with any circle. In the first case, given a point $\left(a,r+\sqrt{r^2-\left(\frac{d_1}{2}\right)^2}\right)$, then we know the center of a circle is $\left(a,\sqrt{r^2-\left(\frac{d_1}{2}\right)^2}\right)$. If $l_3$ neither crosses nor come in contact with any circle, we know the situation is in case2. Letting $l_4:y=r-\sqrt{r^2-\left(\frac{d_1}{2}\right)^2}$, then you do get the point $\left(b,r-\sqrt{r^2-\left(\frac{d_1}{2}\right)^2}\right)$, so we know the center of a circle is $\left(b,-\sqrt{r^2-\left(\frac{d_1}{2}\right)^2}\right)$.

In the case of $l_1\not \parallel l_2$, I use Blue's idea(see comments below). We can move everything such that the intersection comes at the origin and that one of bisectors of two angles between two lines comes x-axis. Without loss of generality, suppose that $l_1:y=mx, l_2:y=-mx$. Letting $l_3:y=0$, we have three different cases: Case$1$ is that the origin is outside of a circle. Case$2$ is that the origin is on a circle. Case$3$ is that the origin is inside of a circle.

Case$1$: Suppose that the points $A, D$ are on the $l_1$ and $C, F$ on $l_2$ and $B, E$ on $l_3$ and that each of $A, B, C$ is nearer to the origin than the other. Letting $OA=a, OB=b, OC=c$, we get the following from the power of a point theorem. $$a(a+d_1 )=c(c+d_2 ), a(a+d_1 )=b(b+d_3 )$$ Then, we can represent $b,c$ by $a, d_i$. After a tedious calculation, we know we can represent every coordinates by $a, d_i$. Then, letting $4(a^2+ad_1 )=k$, we know the following has to be satisfied:$$\left(\sqrt{(d_1)^2+k}+\sqrt{(d_2)^2+k}\right){\sqrt{1+m^2}}=2\sqrt{(d_3)^2+k}$$ By $a=\frac{-d_1+\sqrt{(d_1)^2+k}}{2}$, we know we can represent every coordinates and the center of a circle by $k,d_i, m$. Hence, this is a conclusion: If there exists a positive real number $k$ such that $$\left(\sqrt{(d_1)^2+k}+\sqrt{(d_2)^2+k}\right){\sqrt{1+m^2}}=2\sqrt{(d_3)^2+k},$$(you can get this from $j$ below represented in two ways)and the radius is less than or equals $1$, then the center of a circle can be found: $$(i,j)=\left(b+\frac{d_3}{2},\frac{2b+d_3-(2c+d_2)\sqrt{1+m^2}}{2m}\right)$$ with $$b=\frac{-d_3+\sqrt{{d_3}^2+k}}{2}, c=\frac{-d_2+\sqrt{{d_2}^2+k}}{2}.$$ And the radius is $$r=\sqrt{(b-i)^2+j^2}.$$However, the radius and the coordinates of a center are so complicated that I can't calculate them any more. In the Case$2$ and Case$3$, alomost same argument can be done. Case$2$ is easier than the others. In the case$3$, I have another difficulty: ambiguity in the signs. Strictly speaking, I think we can't decide just one answer. In other words, there remains several possibilities and it's hard to consider the conditions about possibilities.

My answer is like an 'algorism' solution. This is why I'm not satisfied with this answer. Could you give me any hint or another better idea?

Edit: I hope the following two are useful.

I'll use the same alphabets as Blue used.

$1$. Since each of $l_1, l_2$ crosses a circle, considering each of the discriminants of the followings: $$(x-h)^2+(\pm x\tanθ-k)^2=r^2,$$ we get $$\left(-h\mp k\tanθ\right)^2-\left(1+\tan^2θ\right)\left(h^2+k^2-r^2\right)>0$$ $$⇔\ \left(k\mp h\tanθ\right)^2<r^2\left(1+\tan^2θ\right)⇔\ r^2>\frac{(k\mp h\tanθ)^2}{1+\tan^2θ}⇔\ r>\frac{|k\mp h\tanθ|}{\sqrt{1+\tan^2θ}}.$$

Since $r$ is less than or equals to $1$, we get$$\frac{|k\mp h\tanθ|}{\sqrt{1+\tan^2θ}}<1⇔\ |k\mp h\tanθ|<\sqrt{1+\tan^2θ}$$ $$⇔\ h\tanθ-\sqrt{1+\tan^2θ}<k<h\tanθ+\sqrt{1+\tan^2θ},$$$$ -h\tanθ-\sqrt{1+\tan^2θ}<k<-h\tanθ+\sqrt{1+\tan^2θ}.$$ This shows the region in which the center of a circle can be exist.

$2$. Considering the situation that a circle $(x-h)^2+(y-k)^2=r^2$ passes two points $(0,0), (d,0)$, we get $$h=\frac d2, r=\sqrt{\frac{d^2}{4}+k^2}.$$ Since $r$ is less than or equals to $1$, we get $$-\sqrt{1-\frac{d^2}{4}}\le k\le \sqrt{1-\frac{d^2}{4}}$$ for a real number $d$ which satisfies $0\le d\le 2$.

Since we've already known that the lengths of the chords of a circle by two lines $l_1, l_2$ are $d_1, d_2$$(2>d_1\ge d_2>0)$respectively, by the argument above, we know that the center of a circle does exist in a parallelogram-shaped region whose center is the origin: the length of its edges are $2\sqrt{1-\frac{{d_1}^2}{4}}, 2\sqrt{1-\frac{{d_2}^2}{4}}$.

I've tried to find a new line $l_3$ instead of $x=0$, but I'm facing difficulty.

Edit $2$: I've got the following which would be useful.

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 to the radius of a circle which crosses $l_2$. Note that $d$ must satisfy the following because of what I've already written above:$$0\le d\le \sqrt{1-\frac{{d_1}^2}{4}}.$$

Then, Letting each of the intersections of $l_{1,d-}, l_{2,D+}$ and $l_{1,d+}, l_{2,D+}$ and $l_{1,d+}, l_{2,D-}$ and $l_{1,d-}, l_{2,D-}$ 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 as followings: $$C_{-+}:\left(x-\frac{d+D}{2\sinθ}\right)^2+\left(y-\frac{D-d}{2\cosθ}\right)^2=d^2+\frac{{d_1}^2}{4}$$ $$C_{++}:\left(x-\frac{-d+D}{2\sinθ}\right)^2+\left(y-\frac{d+D}{2\cosθ}\right)^2=d^2+\frac{{d_1}^2}{4}$$ $$C_{+-}:\left(x-\frac{-d-D}{2\sinθ}\right)^2+\left(y-\frac{d-D}{2\cosθ}\right)^2=d^2+\frac{{d_1}^2}{4}$$ $$C_{--}:\left(x-\frac{d-D}{2\sinθ}\right)^2+\left(y-\frac{-d-D}{2\cosθ}\right)^2=d^2+\frac{{d_1}^2}{4}.$$

Let's consider the center of $C_{-+}$, which is $P_{-+}$. Letting $P_{-+}$ be $(x, y)$, then we get $d=x\sinθ-y\cosθ$. Putting it in the equation $2y\cosθ=D-d$, we get $$xy=\frac{{d_1}^2-{d_2}^2}{16\sinθ\cosθ}.$$ Hence, we know that the point $P_{-+}$ is on a hyperbola for any $d$.

I've tried to find a special line which is tangent to $C_{-+}$ for any $d$, but I'm facing difficulty. I expect that we can solve this question in an elegant way. I hope I'm getting closer to it.

Edit$3$:

According to Blue's answer, using $l_3$ Blue wrote, we get $$p=1-\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2\sin\theta}>0$$ if $\theta\ge\ \pi/4$.

Also, using $l_4$ Blue wrote, we get $$p=-1-\frac{\sqrt{1-a^2}-\sqrt{1-b^2}}{2\sin\theta}>0$$ if $0<\sin\theta<\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2}(<\frac{1}{\sqrt2})$.

If $\phi=0,$ we get $$0=16h^4\sin^3θ\cos^2θ+8h^2\sinθ\cos^2θ(a^2 +b^2 −2c^2 −2p^2)+8hp{\cosθ}(a^2−b^2 )-(a^2−b^2)^2\sinθ.$$

Hence, we know that there's exactly one positive root $h$ for any negative $p$.

Then,$$y=p=\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2\cos\theta}-1<0$$ if $0<\theta\le \pi/4.$ $$y=p=\frac{\sqrt{1-a^2}-\sqrt{1-b^2}}{2\cos\theta}+1<0$$ if $0<\cos\theta<\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2}(<\frac{1}{\sqrt2})$.

Hence, if $0<\theta<\alpha$ such that $\sin\alpha=\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2}$, then we can choose the following lines: $$l_3:x=\frac{\sqrt{1-a^2}-\sqrt{1-b^2}}{2\sin\theta}+1,$$ $$l_4:y=\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2\cos\theta}-1.$$

If $\beta<\theta<\pi/2$ such that $\cos\beta=\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2}$, then we can choose the following lines: $$l_3:x=\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2\sin\theta}-1,$$ $$l_4:y=\frac{\sqrt{1-a^2}-\sqrt{1-b^2}}{2\cos\theta}+1.$$

Then, the problem for $\alpha\le \theta\le \beta$ remains unsolved. I hope I'm not mistaken.

Edit$\ 4$:

As I've already written above, if $\phi=0$, we get $$0=16h^4\sin^3\theta\cos^2\theta+8h^2\sin\theta\cos^2\theta\left(a^2+b^2-2c^2-2p^2\right)+8hp\cos\theta(a^2-b^2)-(a^2-b^2)^2\sin\theta\ \ \ \ \ \ \ \cdots(\star\star).$$ Changing $h$ to $-h$ in $(\star\star)$ gives us $$0=16h^4\sin^3\theta\cos^2\theta+8h^2\sin\theta\cos^2\theta\left(a^2+b^2-2c^2-2p^2\right)-8hp\cos\theta(a^2-b^2)-(a^2-b^2)^2\sin\theta\ \ \ \ \ \ \ \cdots(\star\star\star).$$

Since we know that there's exactly one positive root $h$ for any positive $p$ in $(\star\star\star)$, we also know that there's exactly one negative root $h$ for any positive $p$ in $(\star\star)$.

In the following argument, suppose that $$\cos\beta=\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2}, \cos\gamma=\frac{\sqrt{1-a^2}+\sqrt{1-b^2}}{2}.$$

Note that if $a>b$, then $$0<\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2}<\frac{\sqrt{1-a^2}+\sqrt{1-b^2}}{2}<1.$$

By the argument above, I got the following:

When $0<\theta<\gamma$, $$l_3:y=\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2\cos\theta}-1\ (<0),$$ $$l_4:y=\frac{-\sqrt{1-a^2}-\sqrt{1-b^2}}{2\cos\theta}+1\ (>0).$$

When $\gamma<\theta<\beta$, $$l_3:y=\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2\cos\theta}-1\ (<0),$$ $$l_4:y=\frac{\sqrt{1-a^2}+\sqrt{1-b^2}}{2\cos\theta}-1\ (>0).$$

When $\beta<\theta<\pi/2$, $$l_3:y=\frac{\sqrt{1-a^2}-\sqrt{1-b^2}}{2\cos\theta}+1\ (<0),$$ $$l_4:y=\frac{\sqrt{1-a^2}+\sqrt{1-b^2}}{2\cos\theta}-1\ (>0).$$

In each case, if $l_3$ crosses an invisible circle at two points, then $(\star\star)$ has exactly one positive root $h$. If $l_3$ neither crosses nor comes in contact with any circle, then $l_4$ gives us $c$, so we'll get exactly one negative root $h$ in $(\star\star)$.

Is this correct? Did I miss anything? I'm afraid I might be mistaken.

Edit $5$:

I got the following:

Letting $\sin\epsilon=\frac{\sqrt{1-a^2}}{2}, \cos\omega=\frac{\sqrt{1-a^2}}{2}$, note that $0<\epsilon<\omega<\pi/2$ for any $a$.

(i) When $0<\theta<\omega$, let's take $l_3$ to be the line: $$y=\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2\cos\theta}+1.$$ Note that this $l_3$ does cross any one of upper-right sub-family of circles because of the following: $$\frac{\sqrt{1-a^2}+\sqrt{1-b^2}}{2\cos\theta}-1<\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2\cos\theta}+1 ⇔ \cos\theta>\frac{\sqrt{1-a^2}}{2}.$$ If we get $c$, we can take a positive root (there's exactly one positive root) in $(\star\star)$ as $h$, because this $l_3$ never crosses the lower-left sub-family of circles: this is because for any $d_1>d_2$, $$\frac{\sqrt{1-a^2}-\sqrt{1-b^2}}{2\cos\theta}+1<\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2\cos\theta}+1.$$

Then, let's calculate $k=\frac{a^2-b^2}{4h\cos\theta\sin\theta}$.

If we don't get $c$, then let's take $l_4$ to be the line: $$y=\frac{\sqrt{1-a^2}-\sqrt{1-b^2}}{2\cos\theta}-1.$$ Note that this $l_4$ does cross any one of lower-left sub-family of circles because of the following: $$\frac{-\sqrt{1-a^2}-\sqrt{1-b^2}}{2\cos\theta}+1>\frac{\sqrt{1-a^2}-\sqrt{1-b^2}}{2\cos\theta}-1 ⇔ \cos\theta>\frac{\sqrt{1-a^2}}{2}.$$

Since we do get $c$, we can calculate $k$.

(ii) When $\omega<\theta<\pi/2$, let's take $l_3$ to be the line: $$x=\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2\sin\theta}+1.$$ Note that this $l_3$ does cross any one of upper-right sub-family of circles because of the following: $$\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2\sin\theta}+1>\frac{\sqrt{1-a^2}+\sqrt{1-b^2}}{2\sin\theta}-1 ⇔ \sin\theta>\frac{\sqrt{1-a^2}}{2}.$$ If we get $c$, we can take a positive root (there's exactly one positive root) in $(\star\star)$ as $h$, because this $l_3$ never crosses the lower-left sub-family of circles: this is because for any $d_1>d_2$, $$\frac{\sqrt{1-b^2}-\sqrt{1-a^2}}{2\sin\theta}+1>\frac{\sqrt{1-a^2}-\sqrt{1-b^2}}{2\sin\theta}+1.$$

Then, let's calculate $k=\frac{a^2-b^2}{4h\cos\theta\sin\theta}$.

If we don't get $c$, then let's take $l_4$ to be the line: $$x=\frac{\sqrt{1-a^2}-\sqrt{1-b^2}}{2\sin\theta}-1.$$ Note that this $l_4$ does cross any one of lower-left sub-family of circles because of the following: $$\frac{\sqrt{1-a^2}-\sqrt{1-b^2}}{2\sin\theta}-1<-\frac{\sqrt{1-a^2}-\sqrt{1-b^2}}{2\sin\theta}+1 ⇔ \sin\theta>\frac{\sqrt{1-a^2}}{2}.$$

Since we do get $c$, we can calculate $k$.

I hope I'm not mistaken.

Edit $6$: Just an idea.

I guess that there exists a line (let's call this $L$) which satisfies the following several conditions. If exists, it can be taken as $l_3$.

1. If we rotate $l_1$ about the point $(\frac{\sqrt{a^2-b^2}}{4\sin\theta}, \frac{\sqrt{a^2-b^2}}{4\cos\theta})$ by some angle, then we'll get $L$. Note that this point is the center of the smallest circle of the upper-right sub-family.

2. $L$ crosses every circle of the upper-right sub-family.

3. $L$ never crosses any circle of the lower-left sub-family.

4. The length of each chord cut by $L$ is different from each other.

It might be difficult to give an actual example as $L$, but it is likely that there would exist such $L$.

Solution 3:

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.