Prove that point is on the perimeter of circle I have a construction as the one in the image below.
How would you prove that the point $I$ is on the perimeter of the circle $C_4$

Here is the exact definition for the construction of the image
Let $C_1$ be a circle with center $O_1$ and radius $1$
Let $C_2$ be a circle tangent to $C_1$ with center $O_2$ and radius $2$
Let $\lambda$ be a line which is tangent to both $C_1$ and $C_2$
Let $C_3$ be a circle tangent to $C_1$, $C_2$ and $\lambda$ with center $O_3$
Let $\kappa$ be a line which goes through the point $O_2$ and is perpendicular to the line $\lambda$
Let $O_4$ be the point of intersection between the lines $\lambda$ and $\kappa$
Let $C_4$ be a circle with center $O_4$ and radius $2$
Let $\rho$ be a line which goes through both $O_2$ and $O_3$
Let $I$ be the point of intersection between the lines $\rho$ and $\lambda$
The circle $C_3$ has the radius $6-4\sqrt2$ but please avoid using this fact in the proof.

My attempts
I tried adding different geometrical constructions, like a square with corner points $O_2,O_4,I$, I also noticed that this is equivalent with the angle $O_4O_2O_3$ being $45^\circ$
However none of the things i tried really leads to a solution.

Context
I'm practicing for a math competition, and I came across the problem of finding the radius of the circle $C_3$ first I ended up here and after assuming it actually was on the circle, I came to the correct result. I'm interested if anyone here could complete my solution.

Clearly we need $\angle O_2IO_4=45^\circ$.
Let $\lambda$ cut $C_1$ at $P$. Let $PQ$ be a diameter of $C_1$. Then $PQO_2O_4$ is a rectangle.
Let $\lambda$ cut $C_3$ at $R$. Now by homothety, $QR$ passes through the tangency point $X$ of $C_1$ and $C_3$.
Note that $\angle PXQ=\angle RPQ=90^\circ$, so $\triangle PXQ\sim\triangle RPQ$. Thus the power of $Q$ with respect to $C_3$ is
$$QX\times QR=QP^2=4.$$
Similarly, if $Y$ is the tangency point of $C_1$ and $C_2$, then $QO_4$ passes through $Y$, and the power of $Q$ with respect to $C_2$ is
$$QY\times QO_4=QP^2=4.$$
Therefore $Q$ lies on the radical axis of $C_2$ and $C_3$.
Let $Z$ be the tangency point of $C_2$ and $C_3$; then the radical axis of $C_2$ and $C_3$ is just the common tangent of $C_2$ and $C_3$ at $Z$. Thus $QZ$ is tangent to $C_2$, and we have
$$QZ^2=4\Rightarrow QZ=2=ZO_2.$$
Thus $QZO_2$ is a right isosceles triangle. By symmetry, $O_2,Z,O_3,I$ lie on a straight line. Hence
$$\angle O_2IO_4=\angle ZO_2Q=45^\circ,$$
as desired.
