1. ## Constructing Circles

I have this problem here which I can't seem to solve.
Here it is:

From the picture above, let's say that the small circle is Circle A, and the big circle is Circle B (the two cross you see are their centers).
I have to construct 5 different circles that is touching Circle A externally and Circle B internally. (using only a compass and a ruler)
I've already constructed two, (the ones where the centers of all three circles lie on a single line) but can't seem to draw the ones that aren't all on a single line.

2. Draw the line joining the two center of the circles. Draw the two circles as you have mentioned. Let R be the radius of the bigger circle. Let R1, R2 and R3 be radii of the three circles in a line from left to right. Draw a perpendicular line to the line joining the centers through the center of the middle circle.
The centers of the remaining two circles must line on the perpendicular line.
Centers of middle circle, right side circle and new bottom circle form a right angled triangle. So
(R2^2 + R4^2) + (R2^2 + R3^2) = (R3^2+ R4^2)
Solving this you can find R4 = R5.

3. I kinda don't understand what you're saying.
Can you show me a picture of it? or put an easier description? I'm a 9th grader by the way.

4. Originally Posted by lanierms
I kinda don't understand what you're saying.
Can you show me a picture of it? or put an easier description? I'm a 9th grader by the way.
For better explanation, you should have given the dimensions of the circles.
Let the diameter of the bigger circle A is 30 cm, and that of smaller circle B is 8 cm. It is at a distance 4 cm from the bigger circle. Draw a circle C with diameter 4 cm and another circle D of diameter ( 30 - 12) = 18 cm. Centers of these circles are in a straight line. Draw a perpendicular line to this straight line through the center of circle B.
Draw a circle E, such that it touches the circles A, B and D. Let its radius be R. Its center lies on the perpendicular line through the center of the circle B.
Centers of the circles B, D and E form a right angled triangle.
So (4 + 9)^2 + (4 + R)^2 = ( 9+R)^2
Solve his equation and find R.
Similarly you can the another circle F on the other side of circle E.

5. Now I get it. Thank you very much. I didn't realize the last equation was the Pythagorean Theorem.
Yeah, but one problem here.
If I do what you told me to do, then the circle E's circumference goes out of the Circle A.
Circle E is only supposed to touch Circle A internally.
Any solution to that?

6. Originally Posted by lanierms
.
If I do what you told me to do, then the circle E's circumference goes out of the Circle A.
Circle E is only supposed to touch Circle A internally.
Any solution to that?
OK.
Now find the distance d between the centers of circle B and D. Find the mid point of this distance. Taking this point as the center and d/2 as the radius draw a circle. Draw a radius from center of A and the center of C. Point of intersection of these radii on the above circle will be the center of the required circle. Distance between the point of intersection and the circumference of the circle A is the required radius.

7. Yeah, but that doesn't work either. It's not the required radius, as the two lines don't seem to have equal length unless it's in a specific position. I'm looking for a way that applies to all possible situations.

8. Originally Posted by lanierms
I have this problem here which I can't seem to solve.
Here it is:

From the picture above, let's say that the small circle is Circle A, and the big circle is Circle B (the two cross you see are their centers).
I have to construct 5 different circles that is touching Circle A externally and Circle B internally. (using only a compass and a ruler)
I've already constructed two, (the ones where the centers of all three circles lie on a single line) but can't seem to draw the ones that aren't all on a single line.
Hi lanierms,

you could locate the centres of the circles you need with the aid of
the attached diagram.
It requires some "tweaking" to get the centres, but it may be sufficient,
unless you require more detailed construction.

Choose a point of contact "a" with the blue circle.
Then draw the green lines from the centres of the blue and black circles.
Adjust the line from the centre of the black circle until you measure

$\displaystyle |ac|=|cb|$

This locates the centre of a circle that is tangential to the blue and black circles, since it will touch both at right-angles to their centrelines.
The point of intersection of the green lines is the centre of such a circle.

9. Thank you very much. Although none of the replies are the ones i'm looking for, I think it's time to give up on this.

10. Originally Posted by lanierms
Thank you very much. Although none of the replies are the ones i'm looking for, I think it's time to give up on this.

Why should you give up?
Can't you use the method which I have described in the thread Drawing lines of equal length?

Try this one.
Let O1 and O2 be the centers of given circles of radii R1 and R2. Let d be the distance between O1and O2 and O be the mid point of O1 and O2. Draw a line joining O1 O2 to meet the circumference of the larger circle.
Now you have to draw an ellipse such that its foci will be O1 and O2 and it passes through the centers of two circles drawn between the given circles.
Let D be the distance between the centers of two circles through which ellipse passes.
Now D/2 will be the semi major axis a of the ellipse.
OO = d/2 will be ae, where e is the eccentricity of the ellipse.
Then semi minor axis b = sqrt[ a^2 - (d/2)^2]
Now we want to draw a circle such that it touches the smaller circle externally and larger circle internally.
For that draw line tangent to the smaller circle, perpendicular the line joining the centers of the circles.
The distance between O and this line will be a*cos(θ). Measure this distance. From that find θ. Νow find b*sinθ. Μark these points on the perpendicular line. This point P will be the center of the required circle. Join O1P and O2P and find the radius of the required circle.

12. That looks good! i must examine that.

I think the following is what lanierms wants
as it only requires a compass and ruler.

First, a point of contact on the large circle circumference needs to be chosen.
Then we draw a second circle with the same radius as the inner circle at that point.

If we draw a centreline of the large circle passing through the point of contact chosen,
then the centre of the required circle lies on this centreline
(so that the new circle touches the large circle at only one point, the chosen one).

We need to find the point that is on this line, equidistant from the 2 circles we started with.
This allows us to draw a circle touching both at a single point.
Hence we can construct an isosceles triangle.

Locate where the new centreline touches the outer circumference of the 2nd pink circle.

Draw a line from this point labelled "a" in the attachment, to the centre of the inner circle marked "b".

Bisect this line and draw the perpendicular bisector.

Draw the line [bc].

We now have an isosceles triangle abc.

$\displaystyle |ac|=|cb|$

However, the radii of the pink circles are equal, hence the red arrows have equal lengths.

Hence "c" is the centre of the required circle.

This procedure can be replicated anywhere in the circle to find the centres of
other similar circles.

13. Thank you very much Archie Meade. That was exactly what I was looking for. I guess I shouldn't have given up on this earlier. Thank you so much.
Thank you also sa-ri-ga-ma, for putting your thoughts into this problem.
Can I ask Archie Meade, what program did you use to draw that picture?
Thank you again.

14. Hi lanierns,

it's a program that came with my Macbook laptop, called "Grapher".
It comes with Mac OSX 10.4 and above.

15. Oh well, since I have a Windows PC, I won't get to use that program.
I'll just use Capri II Plus for drawing the solution. Thanks again.

Page 1 of 2 12 Last