Generalization: $\Gamma_i$ is a circle of radius $r_i$ inscribed in an angle of measure $2\alpha$ such that each $\Gamma_i$ is externally tangent to $\Gamma_{i+1}$ and $r_{i+1} < r_i.$ Then the sum of the areas of the circles $\Gamma_i$ equals the area of a circle of radius $r =\frac {_1}{^2} r_0 (\sqrt{ \sin \alpha} + \sqrt{\text{csc} \alpha}).$ Let $a,b$ the sides of the subject angle with vertex $P$ and denote $X_i$ the tangency point of $\Gamma_i(O_i,r_i)$ with line $a.$ From the similar $\triangle POX_0 \sim \triangle PO_1X_1$ we get $\frac{O_1X_1}{O_0X_0}=\frac{r_1}{r_0}=\frac{PO_1}{PO_0}=\frac{PO_0-O_0O_1}{PO_0}=\frac{PO_0-r_0-r_1}{PO_0}$ Plugging $PO_0=r_0 \cdot \csc \alpha$ yields $\frac{r_1}{r_0}=\frac{\csc \alpha-1}{\csc \alpha+1}=\frac{1-\sin \alpha}{1+\sin \alpha}$ Radii $r_0,r_1,r_2,r_3...$ form a decreasing geometric progression with ratio $k=\frac{1-\sin \alpha}{1+\sin \alpha}$ $\Longrightarrow$ Areas $S_i$ of $\Gamma_i$ form another decreasing geometric progession with ratio $k^2.$ $\sum_{i=0}^{\infty}S_i= \frac{S_0}{1-k^2} = \frac{\pi {r_0}^2}{1-\left ( \frac{1-\sin \alpha}{1+\sin \alpha} \right)^2}= \left ( \frac{1+ \sin \alpha}{2 \sqrt{\sin \alpha}} \right)^2 \cdot \pi {r_0}^2$ Let $\varrho$ be the radius of the circle equivalent to the chain of circles $\Gamma_i.$ Then $\pi \varrho^2=\left ( \frac{1+ \sin \alpha}{2 \sqrt{\sin \alpha}} \right)^2 \cdot \pi {r_0}^2 \ \Longrightarrow \ \varrho=\frac{_1}{^2}r_0 ( \sqrt{\sin \alpha}+\sqrt{\csc \alpha}).$