## Aligning three circle intersection equally

### Question

I am having trouble aligning my objects and I just rely on my eye to have them aligned. What I really want is to have the three "red" intersection to be equal in size

I wanted the three circles to intersect at some point but I am not sure if they do intersect equally.

Here's what I did.

I align the top of the the two circle above and have them intersect at some point. And then align the bottom circle and have them intersect the two.

I just rely on my eye to see if they are aligned but I am thinking if illustrator has some tool to do this.

2014/05/09
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5/9/2014 11:35:00 AM

First create an equilateral triangle.

To do so use the Polygon Tool and bring the number of edges down to 3 using the keyboard arrows. Tap the down arrow to reduce the number of sides. Make sure you hold the Shift key down as you drag with the tool. This ensures the triangle is straight.

Then create your circle and align the center of the circle with one of the corners of the triangle. Enabling Smart Guides (`View > Smart Guides`) will help with alignment.

Duplicate the circle and place the second circle so it aligns with another corner of the triangle. Repeat this for the third circle.

You might want to have `Align to pixel grid` unchecked in the Transform Panel for more precision.

If you're unhappy with the size of your circles use the Scale Tool (s). Select one circle, press enter, specify a scale value. Then repeat for the other two circles one at a time.

2014/05/09
9
5/9/2014 11:33:00 AM

This funny answer for those who like math. It addresses the perfect shape where the center of one circle intersects with the circumference of 2 other ones, however from this explanation one can extrapolate how to achieve the equal distances for any given horizontal displacement.

Let's look at this picture:

We placed two lower circles as described - at 1 distance of their radius.

We can clearly see that the horizontal displacement of the upper circle will always be at 1/2 distance between the lower ones, i.e. 1/2 of the radius.

Second, the centers of the circles are at the radius distance.

Now, we have (I know you feel what I want to do):

So, we have a hypotenuse (r in our case), we have a small cathetus (r/2)... and what we left to do is... to calculate the long cathetus, which is naturally the consequence of Pythagorean theorem.

Thus, long cathetus = âˆš (rÂ²-0.5rÂ²).

So, we create one circle and horizontally displace its copy to 1 radius distance.

Next, we displace another copy horizontally to 1/2 of the radius, and vertically accordingly to the calculated value of long cathetus - with minus sign before it:

And that what we get:

Another case for arbitrary horizontal displacement -

Sad to say, if we could displace the circle by the angle and distance of displacement via transform operation we could achieve perfect results, however, Illustrator convert the distance of displacement incorrectly to vertical displacement after the angle is inserted.

2014/05/09