How to embed this circle tangent to the other circles?

Question

I want to construct a circle that would be tangent to the $3$ circles and would have its diameter lie somewhere on the segment $BI$.

$EF$ includes the diameters of the $3$ given circles. $EB=BF$.

What do you think is the best way of doing this?

Answer 1

You can use circular inversion, or just intersect the segment $BI$ with the hyperbola that is the locus of points $P$ for which $PG-PI=BG$. That intersection is the center of the circle you are looking for.

If you are familiar with the Descartes circle theorem, you know in advance the radius $r$ of your circle, since, given that $BG=1$, $$2\left(\frac{1}{4}+\frac{1}{1}+\frac{1}{1}+\frac{1}{r^2}\right)=\left(-\frac{1}{2}+\frac{1}{1}+\frac{1}{1}+\frac{1}{r}\right)^2$$ must hold, hence $r=\frac{2}{3}$.

By inverting the construction with respect to a circle with center $E$ and radius $EB$, we just have to find a circle that is tangent to two parallel lines and an inner tangent circle, then take its circular inverse: