A circle centered at A with radius 10 is externally tangent to a circle centered at B with radius 7. A line that is externally tangent to both circles is drawn, where both circles lie on the same side of the line. This line intersects line AB at C. Find the length BC.

MathProblemSolver101 Mar 18, 2020

#1**+2 **

ok let's start the problem by first envisioning the two circles. First, we have circle a, with a radius of 10, and circle b with a radius of 7. If we connect AB and extend it into a line, we get the line AB. Drawing a line tangent to both circles creates a right triangle triangle, with the line intersecting line AB at C. Now, we have a big right trianlge with a right angle located at the place where the radius of circle A intersects the tangent line(remember the definition of tangent!) We can then form another right triangle by drawing the radius of circle B to intersect the tangent line. This gives us two similar right triangles(by AA similarity), with the ratio of the big triangle to small triangle being 10/7(the ratio of the radii). With this information in hand, we can solve for the length of the hypotenuse of the large triangle by using the similarity ratio:

if we name the length BC as x, we get 17+x : x = 10:7 , or (17+x)/x = (10/7). Cross multipling, we get 10x = 119 + 7x, simplifying to 3x = 119, or x = **119/3**

jfan17 Mar 18, 2020

#2**0 **

A circle centered at A with radius 10 is externally tangent to a circle centered at B with radius 7. A line that is externally tangent to both circles is drawn, where both circles lie on the same side of the line. This line intersects line AB at C. Find the length BC.

**r _{1} = 10 **

**r _{2} = 7**

**10 + 7 = 17 ( hypotenuse )**

**10 - 7 = 3 ( short leg )**

** ∠BCM = ∠ ABN sin(∠ ABN ) = 3 / 17 ∠ ABN = 10.16424862°**

**BC = BM / sin(∠ ABN) = 39.6666667 units **

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Guest Mar 19, 2020