Intersection between Apollonius circles and circles that passes through $\alpha_1$ and $\alpha_2$

Question

Consider two different points $\alpha_1,\alpha_2\in\mathbb{C}$. Prove that every circle that passes through $\alpha_1,\alpha_2$; intersects every circle of $A$ orthogonally in two points, where $A=\{\{ z\in\mathbb{C}: \frac{|z-\alpha_1|}{|z-\alpha_2|}=r\}: r\in \mathbb{R}^+ \}$

Answer 1

Is it allowed to use inversions or the theory of pencils of circles? Or has it to be proved using just analytic geometry in the complex plane? The following is a prove that assumes one already knows the theory of pencils of circles. Otherwise I will post a different solution using just complex analytic geometry.

The set
$$A=\{\{ z\in\mathbb{C}: \frac{|z-\alpha_1|}{|z-\alpha_2|}=k\}: k\in \mathbb{R}^+ \})$$ is a pencil of circles. In fact from $$\frac{|z-\alpha_1|}{|z-\alpha_2|}=k$$ we have $$ |z-\alpha_1| = k |z-\alpha_2| $$ $$ (z-\alpha_1)( \overline{z}-\overline{\alpha_1}) = k^2 (z-\alpha_2)( \overline{z}-\overline{\alpha_2}) $$ $$ (z-\alpha_1)( \overline{z}-\overline{\alpha_1}) - k^2 (z-\alpha_2)( \overline{z}-\overline{\alpha_2}) = 0 $$

It is now clear that this is a nonintersecting pencil of circles. The two circles of zero radius
$$ (z-\alpha_1)( \overline{z}-\overline{\alpha_1}) = 0 $$ and $$ (z-\alpha_2)( \overline{z}-\overline{\alpha_2}) = 0 $$ are its limiting points and correspond to the values $k=0$ (point $\alpha_1$) and $k=\infty$ (point $\alpha_2$). The radical axis is the perpendicular bisector of the segment $ \alpha_1\alpha_2$ defined by the two point-circles and it corresponds to $k=1$.
The locus of the centers of these circles is the line $ \alpha_1\alpha_2$. Apart from the radical axis, every circle in this pencil contains the point $ \alpha_1 $ or $ \alpha_2 $.

The set B of all circles that pass through $\alpha_1$ and $\alpha_2$ is a pencil too, it is an intersecting circles pencil and its radical axis is the line $\alpha_1\alpha_2$.

From elementary geometry it is now immediate to conclude that every circle in pencil A intersects every circle in pencil B, because every circle in pencil A contains the point $\alpha_1$ or $\alpha_2$, so it must intersect every circle passing through $\alpha_1$ and $\alpha_2$.

We are done: the line of centers of pencil A is the radical axis of pencil B, so all the circles of pencil A are orthogonal to all circles of pencil B. The other way round: the line of centers of pencil B is the radical axis of pencil A.

Answer 2

The second solution to the question uses just analytic geometry in the complex plane. Is is possibile to proceed using two generic points $\alpha_1,\alpha_2\in\mathbb{C}$ as previously, but it is easier to choose two points on the real axis, let's say $\alpha_1=-d$ and $\alpha_2=d$, $d\in\mathbb{R}$. Obviously it is not restrictive, it is like changing reference frame. We have: $$ |z+d|=k|z-d|$$ $$ (z+d)(\overline{z}+d)= k^2(z-d)( \overline{z}-d) $$ $$ z\overline{z} +dz+d\overline{z}+d^2 = k^2(z\overline{z} -dz-d\overline{z}+d^2)$$ and finally $$ z\overline{z} - \frac{(k^2+1)d}{(k^2-1)}z- \frac{(k^2+1)d}{(k^2-1)}\overline{z}+d^2 = 0$$ This is the equation of the Apollonius circle that passes through $\alpha_1$ and $\alpha_2$ because it is known, and easy to proof, that the general equation of a circle is of the form $z\overline{z}+az+\overline{a}\overline{z}+b=0$ with $b\in \mathbb{R}$. The center is $-\overline{a}$ and the square of the radius is $a\overline{a}-b$. In our case, the center is $$z_A=\frac{(k^2+1)d}{(k^2-1)} $$ and it's radius squared is $$r_A^2=(\frac{(k^2+1)d}{(k^2-1)})^2-d^2$$.

Now let's find centers and radii of the circles that pass through $\alpha_1$ and $\alpha_2$. Their centers lie on the perpendicular bisector of the segment $α_1α_2$, that is the $y$ axis. A generic point on the $y$ axis, that is the center of one of the circles, is $$z_B=it, \; t\in\mathbb{R}$$ and its distance squared from $\alpha_1$ (or $\alpha_2$), that is the square of radius of the circle, is $$r_B^2=(d-it)\overline{(d-it)}=(d-it)(d+it) = d^2+t^2$$

The square of the distance between the centers is $$d_{AB}^2=(z_A-z_B)(\overline{z_A - z_B})= (\frac{(k^2+1)d}{(k^2-1)})^2+t^2$$

Now we remember from elementary geometry that intersecting circles are orthogonal if the sum of their radii squared equals the square of the distance between their centers, that is if $$r_A^2+r_B^2=d_{AB}^2$$ and indeed it is immediate to verify that this is true.