What is the area enclosed by the two lines tangent to two tangent circles of different diameter?

Question

Suppose one circle has a radius of 2, and the other's is 3. They touch at one point. Two lines, both touching each circle at one point, enclose an area that I'm not sure how to find. Picture leaning boards against the sides of a snowman. The solution may just require basic geometry, or this may be calculus stuff. If it's the former, just color me stupid, because I can't see it.

I'm looking for anything to get me a handhold. The area of the entire figure, area of any segment, the angle the tangent lines (when extended) make with each other, the perimeter of the figure, the length of any segment of either line... I know I'm grasping at straws, but any insight at all will help me.

Answer 1

The simple solution i had in mind was based on an incorrect assumption... i will add a diagram in a minutes... but the steps are fairly simple.

1 Find the general equation for a tangent line to the big circle on the upper right hand quad. Do the same for the smaller circle.

2 Set equal and solve for the same line.

3 find the x,y pairs on each circle that give you these

4 find the distance 'h' between these pairs

5 find the area of the 2 quadrilaterals which is h*(r1+r2) (thats including both)

6 Given either x,y pair find theta such that y/x = tan theta (at least my belief is that the angle is the same for both circles... you can check to confirm)

7 Find the area of that region intersecting with:
big circle pi*r^2 * [theta/pi]
small circle pi*r^2 * [(pi-theta)/pi]

8 Subtract those from the area of the quadrilateral...

diagram coming... diagram there

Answer 2

Okay, I think I can do it using geometry, some algebra, and possibly some simple trig. No calculus.

One problem is that I'm not quite sure just what area you want to find, but I've sketched it out, and we'll take it from there.

Before we start, there's a theorem to put in the back of our heads: "The measure of an angle formed by two tangents that intersect in the exterior of a circle is one-half the difference of the measures of the intercepted arcs."

I don't know if we'll use that or not, but we ought to be aware of it.

We'll use your example of circles with radii 2 and 3. Start by drawing a long horizontal line (the x axis). Somewhere in the middle of the page, draw a circle of radius 2, tangent to the x-axis. Label the center P and the point of tangency A. To the right of that circle, draw another, tangent to the first and to the x-axis, of radius 3. Label the center Q and the point of tangency B.

Draw PQ connecting the centers, and extend PQ to where it intersects the x-axis. Label that intersection O; draw the y-axis through O, and note that O has the coordinates (0,0).

Draw PA and QB, both perpendicular to the x-axis and parallel to each other. We want to find the distance AB, and we can do so using the Pythagorean Theorem. We know PQ=5, PA=2, and QB=3, so AB^2 = PQ^2 - (QB-PA)^2 = 25 - 1 = 24. Therefore, AB = sqrt(24) = 2 sqrt(6).

The quadrilateral APQB is a trapezoid. (PA and QB are parallel.) AB is the altitude, since AB is perpendicular to both PA and QB.

The area of trapezoid APQB is (1/2)(AB)(PA+QB) = (1/2)(2 sqrt(6))(2+3) = 5 sqrt(6).

I'll stop there, because I don't know what you want to find. To continue, I think you can find the coordinates of P and Q, and the distance OA. You can also get the measures of angles POA and OPA, and you note that triangles OPA and OQB are similar.

In other words, there's a lot you can do now if you want to.

This should've helped.

Answer 3

Yes, geometry should work.

Using anomaly method, but height of trapezium should be the length of the tangent between the touching points with the circles, not on the line passing the centres of the circles.

Quite hard to explaine without diagram, but the idea is to find the trapezium height, let it be AB.
First, connect a line from the touching point of 2 circles (let it be C, and next time label some points to help :)) to the AB at D, such that at C, this line is tangent to the circles. Thus, Let centres of circls be E and F. There are 2 quadrilateral, ADCE and BDCF.
Angles EAD, ECD, FBD, and FCD are all 90 degrees.
angle CFB + angle CDB = 180 = angle CDB + angle CDA
thus, angle CFB = angle CDA, and also, angle CEA = angle CFB.
So, these are similar quadrilaterals.
Let x be the length of CD, then by similarity ratio, CD/AE = BF/CD,
thus x = sqrt(AExBF) = sqrt(product of the 2 radius).
But CD = AD = DB,
thus height of trapezium ABFE, AB = 2 x sqrt(product of the 2 radius)

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Thus, an application of the above is for finding the geometric mean of two numbers a and b by measurements. The steps are:
a. draw a straight line
b. on it, mark 3 points A,B, C such that lengths of AB and BC are a and b respectively.
c. construct 2 circles of radius a and b with centres at A and C respectively.
d. Draw a straight line that just touches the two circles. Let the touching points be D and E.
e. From point C, draw a straight line to DE at F where CF is perpendicular to EF.
f. Measure the length of CF to get the geometric mean of a and b, i.e. length of CF = sqrt(ab).

Answer 4

I am not sure how do I understand your question. But here the solution as far as i can help.

Find the area of imaginary trapezium that created by the tangent line, the tangent line are the side line of the trapezium and the diameter of both circles act as top and bottom line .
r1=2, r2=3, h = 2+3=5
A = 1/2(r2+r1) x h
= 12.5

Next step find the area of half of each circle
A2= 1/2pi X j^2
A3 = 1/2pi X j^2

The you take below calculation to get the answer:

= A - A2 - A3

taraaaa...

Answer 5

Later versions have a 'tangent' feature in 'o snap'. Draw a radius and tangent to the first circle. Then rotate it about the center of the first till it's tangent to the second.

Answer 6

Im thinking its a pool ball. But i really dont know cause i got lost on that first radius.

Answer 7

The question is not clear as you have stated it. Is that precisely how it is worded? What area are you referring to?