To what point or curve does the inversion $I_{0,5}$ transform the circle centered at $(15,0)$ with radius $13?$
$I_{0,5}$ means inverting on a circle of radius $5$ centered at $(0,0).$
I was looking at this question and answer as reference:
The same question but with $I_{0,4}$ and the circle centered at $(0,2)$ with a radius of $1.$
If $q$ is the circle centered at $(0, 2)$ with a radius of $1$ unit, then $I_{0,4}$ is a circle bisected by the $y$-axis. The intersects of $q$ with the $y$-axis are $(0, 1)$ and $(0, 3)$ and these are transformed by to the points $(0, 16/3)$ and $(0, 16).$
So using the idea and applying it to my question:
Then this circle is bisected by the $x$-axis where the intersects are $(2,0)$ and $(28,0)$ but now I'm having a hard time finding out the transformations.
I know by definition in the unit circle:
the transformation occurs like this:
$t: (x,y) \to (\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})$
but this is for radius $1.$
So if anyone can explain how to get the transformation points or in the question I was looking as reference, how did they get $(0,16/3)$ and $(0,16),$ that would help a lot!
i am going to assume that $$I_{0, 5} \text{ means inverting on a circle of radius $5$ centered at $(0,0).$ }$$
the circle centered at $$(15, 0) \text{ and radius $ 13$ has $x$- intercepts at } A = (2,0), B = (28,0)$$ let the images of $A,B$ are $A'=(a', 0), B'=(b', 0)$ respectively. then $$a' \times 2 = 5^2, b' \times 28 = 25 \to a' = \frac{25} 2, b' = \frac{25}{28}$$ the image of the circle cecentered at $(15, 0)$ and of radius $ 13$ is a circle centered at $(\frac{375}{56},0)$ and of radius $\frac{325}{56}.$