Find a hyperbolic rigid motion to transform a semicircle centered on the x axis to another semicircle

Question

I need to find a hyperbolic rigid motion that transforms a semicircle centered at (0,0) of radius 2 to the semicircle centered at (6,0) 0f radius 4. My observations: since a semicircle goes to a semicircle, the center of inversion is not on the original semicircle; since point(2,0)is both on the original semicircle and on the image, it is a "fixed" point of the transformation, so (2,0) is on the circle of inversion. If we call O the center of inversion, k the radius of inversion and Q the point(2,0) then OQ*OQ'=k^2 becomes (OQ)^2=k^2-->OQ=k; and Q is an endpoint of the circle of inversion. I have no idea how to move forward from here. Any hints would be greatly appreciated

Answer 1

Call $c_1$ the circle centered at $(0,0)$ and of radius $2$ and $c_2$ the circle centered at $(6,0)$ and of radius $4$. Take a circle $c_0$ with (Euclidean) center $\big(- 6, \,0\big)$ and (Euclidean) radius $8$ and take a line $l_0$ with equation $x = 6$, i.e. a vertical line, orthogonal to the $x$ axis, passing through the center $(6,0)$ of the second circle $c_2$. Then the inversion $J_0$ with respect to circle $c_0$ maps $c_1$ to $c_2$. After that, the reflection $R_0$ in line $l_0$ maps $c_2$ to $c_2$, simply reflecting it. Then the composition $T = R_0 \circ J_0$ is the hyperbolic translation that maps $c_1$ to $c_2$.

If $z = x + i \, y$ then the formula for $J_0$ is $$J_0(z) = \frac{64}{\bar{z} + 6} - 6$$ and the one for $I_0$ is $$I_0(z) = 12 - \bar{z}$$ The composition is $$T(z) = I_0 \circ J_0 (z) = 12 - \left(\frac{64}{{z} + 6} - 6\right) = 18 - \frac{64}{{z} + 6} = \frac{18z+44}{{z} + 6} $$

Is this what you are asking about?

Answer 2

Your semi- circle centered at $(0,0) $ intersect the x-axis at $ (-2, 0 ) \text{ and } (2, 0 )$

while your semi- circle centered at $(6,0) $ intersect the x-axis at $ (2, 0 ) \text{ and } (10, 0 )$

So your semicircles have a common point at $(2,0)$ To inverse this point $(2,0)$ to itself it has to be on the circle on inversion.

Secondly the point $(10,0)$ has to be mapped to $(-2,0)$

let the center of your inversion circle be $(x,0)$ then the product of the distances from $(x,0)$ to $(-2,0)$ and $(x,0)$ to $(10,0)$ is equal the square of the radius is the square of the distances from $(x,0)$ to $(10,0)$

$$ (x+2)(x-10) = r^2 = (x-2)^2$$ $$ x^2 -8x-20 = x^2-4x+4$$ $$ -8x-20 = -4x+4 $$ $$ x = -6 $$

so the centre of your inversion circle is at $(-6,0)$ and the radius is $ r=|x-2| = 8 $

But then you are talking about a rigid motion the above is a reflection.

Some authors find reflections rigid motions , other authors don't so do check if reflections are rigid motions according to the definitions you have to use.

Good Luck