Prove that point is on the perimeter of circle

Question

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.

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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.

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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.

Answer 1

An alternative solution that does not require the "a priori" knowledge of $C_3$ radius can be obtained as follows. Take the $k$ line as the $x$-axis and the $\lambda$ line as the $y$-axis. The circle $C_2$ has radius $2$ and its center $O_2$ has coordinates $(2,0)$, so that its equation is $(x-2)^2+y^2=4$.

Now let us determine the equation of circle $C_1$ and the coordinates $(1,z)$ of its center $O_1$. The equation of $C_1$ can be written as $(x-1)^2+(y-z)^2=1$. Since it is tangent to $C_2$, the calculation of the intersections between $C_1$ and $C_2$ must give a single solution. These intersections can be defined by solving the equations of both circles for $y$ and then equalizing the two expressions. The equation of $C_1$ gives $ y= z \pm \sqrt{2x-x^2}$, whereas that of $C_2$ gives $y = \pm \sqrt{4x-x^2} $. Taking the appropriate signs according to the geometric construction (i.e. with the minus sign in the expression obtained by $C_1$, and the plus sign in that obtained by $C_2$), we get the equation $ z-\sqrt{2x-x^2}=\sqrt{4x-x^2}$, which solved for $x$ gives a determinant of $8z^4-z^6$. In order to obtain a single solution for the intersection point, the discriminant has to be zero, so that we can write $8z^4-z^6=0$. Because by construction $ z$ has to be positive, this leads to the solution $z=2 \sqrt{2}$. So the circle $C_1$ has center $O_1$ in $(1, 2 \sqrt{2})$ and its equation is $(x-1)^2+(y-2 \sqrt{2})^2=1$.

We can now determine the coordinates $(r,s)$ of the center $O_3$ of the circle $C_3$. Note that, by construction, the value of $r$ also corresponds to the radius of $C_3$. The distance from $O_3$ to $O_1$ is $1 + r$. The distance from $O_3$ to $O_2$ is $2 + r$. Reminding the coordinates of $O_1$ and $O_2$, we can then use the standard formulas for the calculation of the distance between two points to get the following equations:

$$(1-r)^2+(2 \sqrt{2}-s)^2=(1+r)^2 $$ $$(2-r)^2+s^2=(2+r)^2 $$

The solutions of this system are $r=6 \pm 4 \sqrt{2}$ and $s=4\sqrt{2} \pm 4$, where the plus signs refer to the alternative solution correctly pointed out in the comment and characterized by a relatively "large" circle, whereas the minus signs refer to the solution that we are looking for. So the coordinates of $O_3$ are $(6 - 4 \sqrt{2}, 4\sqrt{2} - 4)$.

Now it is sufficient to observe that, drawing from $O_3$ the perpendicular to the $x$-axis and calling $P$ the intersection with the axis, the triangle $O_3 P O_2$ is a right triangle whose legs are $O_3P=s=4\sqrt{2} - 4$ and $PO_2=2-r=2- (6 - 4 \sqrt{2})=4\sqrt{2} - 4$. Because the two legs are equal in length, the triangle $O_3 P O_2$ is a right isosceles triangle. Then, the angle $\angle O_4 O_2 O_3$ is equal to $\frac {\pi}{4}$, which directly implies that the point $I$ is on the perimeter of the circle $C_4$.

Answer 2

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.