Small Circle Large Circle

[guitarguy]

I am struggling with this Trig problem:

There is a small circle which is tangent to a large circle. The points P and R are outside of both circles. The line segment PR goes through the center of both circles. There is another point Q. The line segment PQ is tangent to both circles. The points P, Q and R form an angle which is outside of both circles with P at the vertex. Call the angle at point P: V. The radius of the small circle is a. The radius of the large circle is b.

Show that Sin V = (b-a)/(a+b)

What I have done:

I drew the diagram and labelled the points, angle and radii.

I drew the radius of the small circle, a, to the point where PQ touches the small circle. I called the distance form this point to the vertex of the angle g.

I drew the radius of the larger circle, b, to the point where PQ touches the circle. I called the distance from the point where a touches the small circle to b touches the larger circle h.

I found that Sin V = a/h and Sin V = b/(g+h)

Therefore, a=hSin V and b= (g+h)Sin V
b=gSin V + hSin V

Substituting b=gSin V + a

b-a= gSin V
Sin V= (b-a)/g

I am not able to prove g=a+b?

I think I have done somthing wrong in the expression for Sin V?

I am not sure the angle at the center of each circle is 90 degrees, but I am confused?

Thank you for any help you can give me.

 

[pappus]

I am struggling with this Trig problem:

There is a small circle which is tangent to a large circle. The points P and R are outside of both circles. The line segment PR goes through the center of both circles. There is another point Q. The line segment PQ is tangent to both circles. The points P, Q and R form an angle which is outside of both circles with P at the vertex. Call the angle at point P: V. The radius of the small circle is a. The radius of the large circle is b.

Show that Sin V = (b-a)/(a+b)

...

1. Probably your sketch looks like the attachment.

2. The radius of a circle and the tangent include a right angle at the tangent point. Thus you are dealing with 2 similar right triangles. Use proportions:

\(\displaystyle \displaystyle{\frac ax = \frac{b}{x+b+a}}\)

Solve for x.

3. Use this term for x to simplify:

\(\displaystyle \displaystyle{\sin(V) = \frac ax}\)

 

[pappus]

I am struggling with this Trig problem:

There is a small circle which is tangent to a large circle. The points P and R are outside of both circles. The line segment PR goes through the center of both circles. There is another point Q. The line segment PQ is tangent to both circles. The points P, Q and R form an angle which is outside of both circles with P at the vertex. Call the angle at point P: V. The radius of the small circle is a. The radius of the large circle is b.

Show that Sin V = (b-a)/(a+b)

...

I've found a much easier way to to show that this equation is true:

Draw a parallel to the tangent through the midpoint of the small circle. Then you get a right triangle (marked in grey) whose hypotenuse has the lengt (a + b) and whose leg opposite V has the length (b - a).

Thus \(\displaystyle \displaystyle{\sin(V) = \frac{b - a}{b + a}}\)

 

[guitarguy]

Thanks Pappus. I hadn't thought to draw a line parallel to the tangent through the center of the smaller circle and perpendicular to the radius of the larger circle.

Makes sense now.

 

[Mrspi]

i've found a much easier way to to show that this equation is true:

Draw a parallel to the tangent through the midpoint of the small circle. Then you get a right triangle (marked in grey) whose hypotenuse has the lengt (a + b) and whose leg opposite v has the length (b - a).

Thus \(\displaystyle \displaystyle{\sin(v) = \frac{b - a}{b + a}}\)

nice!!!!

 

[pappus]

nice!!!!

Thank you very much -

coming from you this reply is nearly an accolade for me!

 

[pappus]

Mrspi had a moment of weakness

... are you sure?

Did you notice that there are 4 (four!) exclamation marks. This extra-portion of appreciation can't be a sign of weakness.


... btw: It seems to me that this thread is going a little bit astray, isn't it?