+23 Externally tangent circles with centers at points A and B have radii of lengths 5 and 3, respectively. A line externally tangent to both circles intersects ray AB at point C, where B is on AC. What is the length BC?
To solve this problem, we can use the fact that the line externally tangent to both circles is perpendicular to the segment connecting the centers of the circles (the common external tangent). Let $O_1$ and $O_2$ be the centers of the circles with radii of lengths $5$ and $3$, respectively [1][2]. Let $D$ be the intersection point of the common external tangent with $AB$. Then, we have $AD=5+3=8$.
Since the line is perpendicular to $O_1O_2$, we have $\triangle O_1O_2D$ is a right triangle with hypotenuse $O_1O_2$ and leg $BC$. Therefore, by the Pythagorean Theorem, we have $$ BC^2 = O_1O_2^2 - O_1D^2 = (5+3)^2 - 5^2 = 3^2 \cdot 4 = 36. $$
Taking the square root of both sides, we get $BC=\sqrt{36}=6$, which is the length we were asked to find.
Therefore, $BC=\boxed{6}$.
