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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?

 Mar 11, 2023
 #1
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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, $BC=\boxed{6}$.

 Mar 11, 2023
 #2
avatar+118617 
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This question appears to be trivial

 

 Mar 12, 2023
 #3
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'B is on AC'  suggests that B lies between A and C.

The tangent should be the one that runs across the tops (or bottoms) of the two circles crossing AB extended to the right of B.

Guest Mar 12, 2023
 #4
avatar+118617 
+1

Yes I think you're right.  Thanks for pointing that out guest  cool

 

Here is the desired pic

 

Use similar triangles to solve for x

 

Melody  Mar 12, 2023
edited by Melody  Mar 12, 2023

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