Touching Circles [Pythagorean Theorem (direct and inverse).]

Two circles with radii \( r \) and \( R \) have centers at points \( O_{1} \) and \( O \) and touch externally at point \( K \). At point \( A \) on the circle with radius \( R \), a tangent is drawn that intersects the circle with radius \( r \) at points \( B \) and \( C \). It is known that \( B C: A B = p \) and the segment \( A C \) intersects the segment \( O_{1} K \). Determine:

a) under what conditions on \( r, R \), and \( p \) such a geometric configuration is possible;

b) the length of the segment \( B C \).
Let $A B=x, B C=p x$. Drop a perpendicular $O_{1} F$ from point $O_{1}$ to line $O A$ and apply the Pythagorean theorem to the right triangle $O_{1} F O$.

## Solution

Let $A B=x$. Then $B C=p x$. Let the line $A C$ intersect the segment $O_{1} K$ at point $N$, and let $M$ be the midpoint of chord $BC$. Then $O_{1} M \perp B C$ and

$$
O_{1} M=\sqrt{O_{1} C^{2}-C M^{2}}=\sqrt{r^{2}-\frac{p^{2} x^{2}}{4}}
$$

Drop a perpendicular $O_{1} F$ from point $O_{1}$ to line $O A$. Then

$$
\begin{gathered}
O F=O A+A F=O A+O_{1} M=R+\sqrt{r^{2}-\frac{p^{2} x^{2}}{4}} \\
O_{1} F=A M=A B+B M=x+\frac{p x}{2}
\end{gathered}
$$

Since $O_{1} O^{2}=O_{1} F^{2}+O F^{2}$, we have the equation

$$
(r+R)^{2}=\left(x+\frac{p x}{2}\right)^{2}+\left(R+\sqrt{r^{2}-\frac{p^{2} x^{2}}{4}}\right)^{2}
$$

After expanding the brackets and combining like terms, we get the equation

$$
\begin{gathered}
R \sqrt{4 r^{2}-p^{2} x^{2}}=2 r R-x^{2}(p+1) \\
4 r^{2} R^{2}-p^{2} x^{2} R^{2}=4 r^{2} R^{2}-4 r R x^{2}(p+1)+x^{4}(p+1)^{2} \\
x^{2}(p+1)^{2}=4 r R(p+1)-p^{2} R^{2}
\end{gathered}
$$

from which

$$
x=\frac{\sqrt{4(p+1) R r-p^{2} R^{2}}}{p+1}
$$

Therefore,

$$
B C=p x=\frac{p}{p+1} \sqrt{4(p+1) R r-p^{2} R^{2}}
$$

Since $4(p+1) R r-p^{2} R^{2}>0$, we have the inequality

$$
\frac{r}{R}>\frac{p^{2}}{4(p+1)}
$$

Consider the case when the line $A C$ passes through point $O_{1}$. Then

$$
\begin{aligned}
& O_{1} A=\sqrt{(R+r)^{2}-R^{2}}=\sqrt{2 r R+r^{2}} \\
& A B=A O_{1}-O_{1} B=\sqrt{2 r R+r^{2}}-r, B C=2 r
\end{aligned}
$$

Thus, in this case,

$$
\frac{B C}{A B}=\frac{2 r}{\sqrt{2 r R+r^{2}}-r}
$$

For the intersection point of lines $A C$ and $O_{1} O$ to lie on the segment $O_{1} K$, it is necessary and sufficient that the inequality

$$
p>\frac{2 r}{\sqrt{2 r R+r^{2}}-r}
$$

holds. Then

$$
\sqrt{2 r R+r^{2}}>r+\frac{2 r}{p}, \sqrt{\frac{2 R}{r}+1}>1+\frac{2}{p}
$$

from which

$$
\frac{R}{r}>\frac{2(p+1)}{p^{2}}, \frac{r}{R}<\frac{p^{2}}{2(p+1)}
$$

Therefore,

$$
\frac{p^{2}}{4(p+1)}<\frac{r}{R}<\frac{p^{2}}{2(p+1)}
$$

## Answer

a) $\frac{p^{2}}{4(p+1)}<\frac{r}{R}<\frac{p^{2}}{2(p+1)}$;

$$
\text { b) } B C=\frac{p}{p+1} \sqrt{4(p+1) R r-p^{2} R^{2}} \text {. }
$$