Denote: $\Gamma(O,R)$ the given circle with center $O$ and radius $R$; $OP=a$. Let be $K$ the point on the circle $\Gamma$ such that $P\in(OK)$. Let be $\Gamma_1(O_1,x), \Gamma_2(O_2,y)$ the circles tangent to $\Gamma$ internally and to each other at point $P$, such that $\angle{O_1PK}=\angle{O_2PO}=\alpha\in[0,90^\circ)$. Denote $\Gamma\cap\Gamma_1=\{M\}; \Gamma\cap\Gamma_2=\{N\}$. $O_1,P,O_2$ are collinear; $O,O_1,M$ are collinear; $O,O_2,N$ are collinear. Results: $\sqrt{(a+x\cos\alpha)^2+(x\sin\alpha)^2}=R-x\Longrightarrow x=\dfrac{R^2-a^2}{2(R+a\cos\alpha)}$. $\sqrt{(y\cos\alpha-a)^2+(y\sin\alpha)^2}=R-y\Longrightarrow y=\dfrac{R^2-a^2}{2(R-a\cos\alpha)}$. Let be $l_1,l_2$ the two external tangents to $\Gamma_1,\Gamma_2$ and $l_1\cap l_2=\{T\}$. $O_2T=\dfrac{y(x+y)}{y-x}$. $PT=O_2T-PO_2=\dfrac{y(x+y)}{y-x}-y=$ $=\dfrac{2xy}{y-x}=\dfrac{R^2-a^2}{2a\cos\alpha}$. Let be $PT_0$ the orthogonal projection of $PT$ on the line $OP$. $PT_0=PT\cos\alpha=\dfrac{R^2-a^2}{2a}=\text{constant}$. Hence, the locus of $T$ is a line $L$ with the properties: $L\perp OP$; $d(P,L)=\dfrac{R^2-a^2}{2a}$. $d(O,L)=d(P,L)+a=\dfrac{R^2+a^2}{2a}$.