Proof that line passing through centre of circle is mapped to a line under inversion transformation

Question

'Here b is an arbitary point on L, while a is the intersection of perpendicular line with q. By virtue of (5), $ <q\tilde{b}\tilde{a}=<qab= \frac{\pi}{2}$, $\tilde{b}$ lies on same line-segment $ q \tilde{a}$ as diameter. Done."

-Page 127 of Tristan Needham's book Visual Complex Analysis

note: (5) is property that triangle $ q \tilde{b} \tilde{a}$ and $ qba$ are similar.

I do not understand how the fact of two angles being 90 degrees and also triangles being similar implies that each point on the line has a unique map onto a point on the circle.

Answer 1

After about a year, I finally got what was going on here. I will write the steps of the proof pointwise.

1. We know that inversion is self inverse. So, if we prove that line gets mapped to circle, that's enough to say the circle goes to the line.

2. Suppose we have two fixed points $A$ and $B$, then the set locus of points $P$ for which $AP$ and $PB$ are perpendicular is a circle:

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$\quad \quad$ Proof:

$\quad \quad$For simplicity, we can consider one of the fixed points as origin.

$ \quad\quad$ Given $A=(0,0)$ , $B=(B_x,B_y)$ , $P=(x,y)$, $AP \cdot BP =0$:

1. $\quad \quad $$AP \cdot BP = 0$ (From Given)
2. $\quad \quad $$ (x,y) \cdot (B_x -x , B_y -y )=0$ (From 1 )
3. $\quad \quad $ $ x^2 +y^2 - xB_x - y B_y = 0 \iff (x - \frac{B_x}{2})^2 + (y- \frac{B_y}{2})^2 = \frac{ B_x^2 + B_y^2}{4}$
4. $\quad \quad$ From knowledge of conic sections, we know that final equation is a circle.

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3. Join the line segment from center of inversion to nearest point of line $l_a$, call this point $a$. Now, choose an arbitrary point $b$ on the line $L$ and join a line segment $l_b$ from center of inversion to this point on the line $L$. Draw a perpendicular from $l_a$ to $l_b$. Also, let $ \tilde{a}$ on $l_a$ and $\tilde{b}$ and $l_b$ be the inverted points of $a$ and $b$ respectively.

4. Through properties of inversion, one can check that $ q \tilde{b} \tilde{a}$ is necessarily ninety degrees (refer). We also have the fact that $b$ was arbitrary, this means we have the case in mainpoint-2 meaning equivalently that we have a circle.

Therefore, circles get mapped to lines and lines get mapped to circles.

Answer 2

Better perhaps to word it this way:

Proof that the circle passing through centre of circle is mapped to a straight line under inversion transformation.

Zero distance point of circle $\tilde{a}\tilde{b} $ from center of inversion maps to $ ab$ at infinite distance on/as a straight line.

As it is, circles not passing through centre of circle are mapped to other circles.

Under inversion transformation due to similar triangles mapping is along the red lines indicated.

Mapping is point to point to start with.

$$ r\rightarrow \dfrac{R^2}{r}$$

for each cartesian coordinate

$$(\tilde{x},\tilde{y})=\dfrac{R^2(x,y)}{x^2+y^2}$$

angles are mapped as their complement, so reversal occurs in their sense/direction later when similar triangles are considered.

$$ qb\cdot q\tilde{b}= R^2$$

$$ qa\cdot q\tilde{a}= R^2$$