Inversion in a circle

Mondo

Junior Member
Joined
Apr 23, 2021
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127
Hello,

I try to understand the inverse of a line in a circle as described here -> https://i2.paste.pics/a96ec4e85e5f25c0080e978056bd1250.png
From my understanding [MATH]k^2 = r*r' [/MATH] hence he defined r=k2r. From this I would write rcos(θ)=k2cos(θ)r but author for some reason wrote this as [MATH]\frac{k^2r'cos\theta}{r'^2}[/MATH] (any idea why?)
The first serious question I have for the equation [MATH]c(X'^2+Y'^2) + 2k^2(gX' + fY') + k^4[/MATH] Author says it was derived by the multiplication of the equation for C(X,Y) by (X2+Y2) but I don't get that result. For example how he got k4 there?
Next, how he derived the formula for the R (radious of the new circle) [MATH]R'^2 = \frac{k^4(g^2+f^2)}{c^2} \cdot \frac{-k^4}{c}[/MATH]Likewise I dont get the center of the newly created circle (the one after the inversion) as (-gk2c,-fk2c) so multiplied by k2c. The k2=rr and I don;t see the linkage with the center. I would rather say that the new center is (-(g+c),-(f+sth))

Thanks for the help.
 
Last edited:
I think I solved most of the questions...will update soon...
 
Beer soaked comment follows.
I think I solved most of the questions...will update soon...
Can you post your work?
Others can learn from it.

P.S. From which book was that page from?
 
jonah2.0 sure, the book I read is Geometry: A Comprehensive Course by Dan Pedoe.

Autor rely on two circle equations - the original and the transformed one and from that stand point its clear. The only unknown for me is how he got k4 in the transformed equation.
 
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