Smallest circle enclosing 3 circles?

Question

If you have three 15 inch circles, what is the smallest diameter of a fourth circle that could fully contain the 3 smaller ones?

Answer 1

Interesting question. Please pardon my iterative approach to an answer.

If ABC is the triangle whose vertices are the centers of the three 15-inch (diameter?) circles, the simplest solution for the smallest diameter of a fourth circle that includes them all could simply be a circle centered on the the longest side of ABC with a diameter equal to the length of that side plus fifteen inches- but we have to look further to see if that is the case.

Next, you have to use the law of cosines to evaluate whether or not the circle around the remaining vertex is within the circle we just constructed. If the line between the midpoint of the longest side and the remaining vertex plus the radius of the small circle is shorter than the radius of our candidate fourth circle, the problem is solved.

Otherwise, we have to write equations for the two lines that are normal to- and bisect- the enclosing sides. The point those lines cross will be the center of a circle (the "circumcircle") that includes all three vertices of ABC.

Applying the geometry of a 30-60-90 triangle to an equilateral triangle with 15-inch sides you can come up with a minimum diameter (unless you allow overlap) of (3+2*sqrt(3))/3 times the small-circle diameter (a little over 32.32 inches), which is the case when all three circles are touching each other.

The more challenging problem is the location and radius of the fourth circle that is TANGENT to three arbitrary circles (See reference 1). The path to that answer is suggested above, but the actual answer is had by evaluating four determinants (equations 4, 6, 7, 8) and the expression for the circumradius (equation 13) of reference 2 below.

That's too much arithmetic for me. Good luck! :)

Answer 2

The smallest circle that could fully contain three 15 inch circles is just over 15 inches - if you stack the other three. If you do not stack them but have them touching each other you would need a circle that is 33 inches in diameter.

Answer 3

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 4

by geometry. if all the 3 circles where position as tangent to each other by constructing a symmetry triangle with each side 15 inch (internal angle each 60 deg). then construct the 3 circles with its center at each tip of the triangle. then construct another 2 line perpendicular to either of side of triangle to get the center point of minimum circumscribe all the 3 circle. using this center point and 1 of the outer tangent point of either 3 circle (dia 15 inch) to construct the circumscribe circle. measure the diameter of this circle and this is the smallest circumscribe circle that contain all 3 circle which has 3 tangent point to the respective circle. try it with an auto cad or hand tools. approximate dia is 32.3205 inch

Answer 5

15 inches plus the thickness of the material that the circle is made of. Stack the first three circles on top of each other.

Answer 6

a 15 inch circle would circumvent all the circles

Answer 7

About 32 inches...

Answer 8

i can answer only if i know the relative position of each 15inch circle.

Answer 9

radius=15(2+root3)/root3

approximately
32.320508075688772935274463415059