Möbius Geometry Given two non intersecting clines and a third that intersects the first two, prove that there is a unique fourth cline perpendicular to the three given ones.
I am thinking if the two non intersecting clines are $y = 0$ and $y = 1$, and the third is $ y = x$. How can the fourth one be perpendicular to those three. 
And actually how can I start proving the statement? I would like to get some hints here. thanks 
 A: Edit: Since writing this, I've come up with a more elegant solution. I'll leave this here in place in case someone has to do actual computations with these beasts, but would consider my other answer to be more useful for the original question.
Möbius coordinates
You can turn each circle into a vector like this:
$$\begin{pmatrix}x\\y\\x^2+y^2-r^2\\1\end{pmatrix}$$
Two circles (with coordinate vectors $a$ and $b$ formulated as above) are orthohgonal iff the following equation in these coordinates holds:
$$2a_1b_1 + 2a_2b_2 - a_3b_4 - a_4b_3 = 0$$
So for three circles, this gives you three orthogonalities, which leads to three equations. You can plug in a vector with three variables and a one in the last component in order to obtain three equations linear in these variables. Then you have the coordinates of the center for the requested circle, and can use these to compute the radius using the third coordinate. The key point here is the fact that in general the linear system of equations has exactly one unique solution.
Not only circles
If your clines are not circles but lines, you'll have to express them differently. A line $ax+by=d$ will result in a vector
$$\begin{pmatrix}a\\b\\2d\\0\end{pmatrix}$$
You can use this for input, plugging it into the same equations as described above. For output, you should use four variables. Three equations in four variables will usually lead you to a one-dimensional solution space. Pick any non-zero solution. If its last coordinate is zero, you have a line. If it is non-zero, scale all coordinates so you have a one in that coordinate, and you have a circle in the above sense.
Sources
I formulated this based on section 3.2 of this paper by Bobenko and Suris. Any mistakes made in the reformulation are likely mine; I tried to keep the syntax focused on the task at hand.
The central property for two circles to be orthogonal is
$$(x_1-x_2)^2 + (y_1-y_2)^2 = r_1^2 + r_2^2$$
which is equivalent to the first equation stated in my answer.
Relation between circles
Your question asks about a specific relation between circles, so you have to check how these situations correspond to the coordinate vectors I just formulated. The symmetric bilinear form used in my expression relates to the cosine of the intersection angle. The absolute value of that cosine will exceed one if the circles do not intersect at all. However, the bilinear form will be the cosine only for specific multiples of $a$ and $b$, and I don't have the correct scale factor just now.
I'm not sure how knowing that two pairs of circles do intersect but the third one does not will guarantee the existence of the final solution, ie. will guarantee that the $r$ you compute will be a real number and not an imaginary one. You'll have to show that your final solution satisfies
$$a_4(a_1^2+a_2^2-a_3)\geq 0$$
A: Here is another solution, more geometric than my previous one, and better suited to handle the specific configuration you are given.
Generic case
For the moment I'll exclude the special where any of your clines merely touch. See the following figure for the basic idea as well as the labels I'll be using.

So you have two circles intersecting a third one, but not intersecting one another. Have a look the four distinct points of intersection. You can compute their cross ratio
$$\lambda:=\operatorname{CR}(A_1,B_1;C_1,D_1)$$
It will be a real number, as all four points lie on a single circle, namely the green one. If you take the points in cyclic order, you will even get
$$1<\lambda<\infty$$
Now you can draw a rectangle $A_2,B_2,C_2,D_2$ whose corners have this same cross ratio. You can choose the edge lengths of that rectangle to be
\begin{align*}
\overline{A_2D_2} = \overline{B_2C_2} &= 1 \\
\overline{A_2B_2} = \overline{C_2D_2} &= \sqrt{\lambda-1} \\
\end{align*}
With the inequality for $\lambda$ you see that the lengths will all be positive finite real numbers. Now you may know that a Möbius transformation is uniquely defined by the images of three points. So you can map three of your intersection points to three corners of the rectangle, say
$$A_1\mapsto A_2,\quad B_1\mapsto B_2,\quad C_1\mapsto C_2$$
The fourth point of intersection, $D_1$ will automatically get mapped to the fourth corner of the rectangle, $D_2$, as a Möbius transformation will preserve cross ratios. What is important is that the mapping will preserve the cyclic order of your points of intersection. So if your original clines didn't intersect, the points were grouped in two pairs, not interleaved. So the same will hold for the rectangle.
So the images of your original clines under this Möbius transformation are the circumcircle of the rectangle as well as two clines which intersect the rectangle in pairs of adjacent corners. For this reason, the centers of all three image clines will lie on a single line. That line (drawn in black) is therefore orthogonal to all three clines. Take that line and apply the inverse transformation, and you have the cline you asked for (black as well).
Special cases
The case of two clines touching in a single point should be handled separately. If one pair of circles touches in a single point, you can choose any isosceles triangle instead of the rectangle:

If both pairs merely touch the central circle, then any cline which intersects the central circle orthogonally in these two points will automatically intersect the other two orthogonally as well. You can either construct such a circle directly, or use the same mapping approach for this case as well. Now you'd want to map the contact points onto a line and the central circle onto the Thales circle for that line. You can achieve that by mapping an arbitrary point on the central circle to an arbitrary point on that thales circle.

Your example
The example you states in your question, using $y=0$, $y=1$ and $y=x$, does not satisfy the requirement, as all three lines intersect at the infinite point. The first two have no other intersection, so they touch in that point. Therefore your situation is combinatorially equivalent to the following:

In this case, the orthogonal cline degenerates to a single point, the one where all three clines meet. Which is the infinite point in your example.
A: I treat the case where the first circle $\gamma_1$ is the unit circle $\partial D$, and the second circle $\gamma_2$ has center $a>0$ and radius $r>0$, whereby $a+r<1$. (The case where the two circles lie apart is similar.)
There is a Moebius transformation mapping $D$ to itself and transforming $\gamma_2$ into a circle $\gamma_2'$ with center at the origin. This transformation has the form
$$z\mapsto T(z)={z-c\over 1-cz}, \qquad c\in\ ]{-1},1[\ ,$$
and $c$ can be determined from the condition $T(a-r)=-T(a+r)$.
The third circle $\gamma_3$ will be mapped onto a circle $\gamma_3'$ with center $q\in\Bbb C$ (resp., onto a line $\ell$). Consider the line through $0$ and $q$ (resp.,  the perpendicular from $0$ to $\ell$). This line is a generalized circle that intersects $\gamma_1'$, $\gamma_2'$, and $\gamma_3'$ orthogonally, and it is obviously the only such circle.
Now transform back to the original disposition.
