# A math sum involving cirlces!

OK there is actually a diagram unfortuanately I can’t put it up so I’ll to explain it as well as I can.

Here goes

There are five circles in a row. The next circle is bigger than the previous one. All five circles touch each other . Two lines act as common tangents to the circle (one below the circles and one above the circles , picturize.) The smalles circle has a radius of 8 and the bigger circle has a radius of 18.

Find the radius of the the middle circle
PS :00000

something like this but only that they are connected to each other with 2 common tangents one below and above them. No more data been given . I tried using similar triangles but to no avail.

See ya.

Your description is pretty imprecise, but I’ll give it a shot.

I assume that there’s one line connecting the centers of all circles, and that any two circles intersect in at most one point.

Furthermore, there are two lines which are simultaneously tangent to each of the five circles. These two lines intersect in a point to the left of the leftmost circle, yes?

I’ll have to look at it a little more, but off the top of my head I think that if there’s a common difference between the radii of the circles (so that the difference between the radii of a circle and the smaller circle to the left is constant), you’d get the right situation.

Yup they do intersect at a point on the left of the smaller circle. A line joining the centre is not shown in the diagram and if u could prove the radii was in A. P or GP then the sum would be solved.

According to me the constant between the radii of the two adjacent circles will not remain same.

Close. There’s a common ratio. It’s hard to describe how to show this, but I’ll scan something in. In the meantime, if the ratio is k, then:[ul]18 = k[sup]4[/sup] × 8[/ul]
and (where x is the answer):[ul]x = k[sup]2[/sup] × 8[/ul]So this gives:[ul]x = sqrt(8 × 18) = 12[/ul]

Have you proved that, or are you just guessing?

My last response was to Jukeball. Achernar snuck in while I was doing a little algebra (didn’t get anywhere).

The radii of the two circles are “a” and “b”. The important thing is that the ratio of the short leg of the triangle (a-b) to the hypotenuse (a+b) is constant. It depends only on theta, the angle between the tangent line and the line through the centers. The fact that there is a tangent line on both sides of the circles guarantees that this angle will be constant. Although it’s not necessary for solving the problem:
[ul]k = (1 + sin(theta)) / (1 - sin(theta))[/ul]I would appreciate your checking my math on this, ultrafilter. I always seem to mess it up at the worst times.

Which line does r(a + b) measure?

r(a + b) is the short leg of the triangle.

What does this mean?

Never mind, I figured that out. It looks good.

Also, k may be more succinctly expressed as (tan(theta) + sec(theta))^2.

Ultrafilter : I thought you meant the differrent radii were in arthimetic progression . I took extreme cases in the sense really big circles and really small circles. The difference in radii would ofcourse not be constant.
And thanks to both of u’ll!

The fact that all five circles have a common tangent places some pretty severe restrictions on what the radii can be (and it also requires that the centers be collinear).

Agreed but the the two tangents are always going to diverging from one another at one end. And as proved by Achernar the radii are in geometric progression so the the difference in radii cannot be the same.

You know what problem solved , end of matter ,end of thread
G’nite

Sorry, it’s the fact that all five circles have two common tangents that makes the centers collinear.