Radius sum of tangent circles

Denote $\mathcal{C}'$ to be a circle which satisfy the condition and $Oz$ to be the angle bisector ray of $\angle xOy$ .

• Observation:

Firstly, since $\mathcal{C}'$ tangent to both $Ox$ and $Oy$ , the center of $\mathcal{C}'$ is on $Oz$ . Furthermore $\mathcal{C}$ and $\mathcal{C}'$ is tangent to each other, so the tangent point of these two circles is on the line that connect their centers, which is again on $Oz$ . Because $\mathcal{C}$ and $Oz$ is known, we can find the points where $\mathcal{C}$ and $\mathcal{C}'$ touch and draw a common tangent line of the two circles through the points. These lines together with $Ox$ , $Oy$ are the three tangent lines needed to characterize and construct $\mathcal{C}'$ .

• Construction of $\mathcal{C}'$ (and a proof of $\mathcal{C}'$ ):

Let $\mathcal{C}$ intersects $Oz$ at $P_1$ and $P_2$ , with $P_1$ on the line segment $OM$ . These two points are where $\mathcal{C}$ and $\mathcal{C}'$ touch. I will detail the construction of $\mathcal{C}'$ which touches $Ox$ , $Oy$ and $\mathcal{C}$ at $P_1$ only, since we apply the same method for $P_2$ :

For the circles that is tangent to $Ox$ , $Oy$ and $\mathcal{C}$ at $P_1$ (Illustrated in the picture), draw a tangent line of $\mathcal{C}$ at $P_1$ , which intersects $Ox$ and $Oy$ respectively at $A_1$ and $B_1$ . The angle bisectors of $\angle OA_1P_1$ and $\angle P_1A_1x$ cross $Oz$ at $N_1$ and $N_2$ respectively. Then the circles $(N_1,N_1P_1)$ and $(N_2,N_2P_1)$ are two possibilities of $\mathcal{C}'$ .

Proof $(N_2,N_2P_1)$ is a possible $\mathcal{C}'$ : Because $N_2$ is on $Oz$ - the angle bisector of $\angle xOy$ - and $\angle P_1A_1x$ , we have $d(N_2,Oy) = d(N_2,Ox) = d(N_2,P_1A) = N_2P_1$ , as $N_2P_1 \perp P_1A$ .

• Find the radii of all possible $\mathcal{C}'$ :

Since $OA_1B_1$ is an equilateral triangle and $(N_1,N_1P_1)$ is its inscribed circle, we have $N_1P_1 = \frac{1}{3}OP_1 = \frac{1}{3} (OM - MP_1) = \frac{1}{3} (3 - R)$ . Meanwhile $OA_1N_2B_1$ can be proved to be a rhombus, giving $N_2P_1 = OP_1 = 3 - R$ .

For the circles that is tangent to $Ox$ , $Oy$ and $\mathcal{C}$ at $P_2$ , following the same construction method mentioned above we get two other possibilities of $\mathcal{C}'$ with radius $\frac{1}{3}(3 + R)$ and $3 + R$ .

• Conclusion & additional notes:

In total, there are four possibilities of $\mathcal{C}'$ and the radius sum of which is $\frac{1}{3}(3 - R) + (3 - R) + \frac{1}{3}(3 + R) + (3 + R) = 8$ .

Note: The condition $R \neq \frac{OM}{2}$ is added because otherwise $\mathcal{C}$ would be tangent to $Ox$ and $Oy$ , resulting in less possibilities for $\mathcal{C}'$ which doesn't sum up to 8.