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Let there be a circle $$C: (x-3)^{2}+y^{2}=1$$ also let there be a line $$e: x=2$$ Lets consider an inversion in respect to circle $C$. The image of $e$ is another circle $$C_{1}: (x-\frac{5}{2})^{2}+y^{2}=\frac{1}{4}$$

Lets consider any point from this circle, and denote it by $t\in[0,2\pi]$ The function $\phi:[0,2\pi]\rightarrow \mathbb{R}\cup\{\infty\}$ is an inversion of the circle to the line.

Calculate the formula for $\phi$

Is this true that for some constants $a,b,c,d$ we have $\phi(t)=\frac{at+b}{ct+d}$?

I also calculated two extreme values: $\phi(\pi)=(2,0)$ and $\phi(0)=\infty$

Regards

mkultra
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1 Answers1

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No your function isn't of the form : $\frac{at+b}{ct+d}$

It uses trigonometric functions :

$$\phi(t)=\frac{\sin(t)}{1-\cos(t)}\tag{1}$$

Relationship (1) is obtained by expressing the alignment of

  • $\Omega (3,0)$ (circle's center),

  • $M(5/2+(1/2)\cos t, (1/2) \sin t)$ (point on the circle),

  • $M'(2,\phi(t))$ (image on the line of point $M$)

by the following determinant equal to $0$ (see here) :

$$\begin{vmatrix}3 & 5/2+(1/2)\cos t & 2\\ 0 & (1/2) \sin t & \phi(t)\\ 1 & 1 & 1 \end{vmatrix}=0 \tag{2}$$

Expanding (2) gives (1).

Jean Marie
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