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Circle Enclosed by Three Circles

```
Date: 08/18/99 at 11:04:15
From: Nathan Sokalski
Subject: How to find the enclosed circle

I want to know how to find the center and radius of a circle that is
enclosed by three other circles that all touch each other at one
point. Example:

Circle 1      Circle 2      Circle 3
Center X      0            .5            .5
Center Y      0            .5           -.5

I know how to find a circle using 3 points, but I don't know how to
find which one point on each circle the enclosed circle will pass

Thank you.
Nathan Sokalski
```

```
Date: 08/18/99 at 13:19:13
From: Doctor Rob
Subject: Re: How to find the enclosed circle

Thanks for writing to Ask Dr. Math.

Let the center of the circle be (h,k). Then you can get equations for
h and k from the fact that the distance from (h,k) to the center of
each of the three given circles, minus the radius of that circle, is
equal to the radius r of the circle sought. This gives you three
equations in the three unknowns h, k, and r, of the second degree.
Solving these three equations simultaneously will give you the answer
you seek. Careful, though, because the process of solving will
probably give you extraneous solutions, too, which you have to discard
after checking.

In the example you give above, the first circle radius you wrote,
(2^2-1)/2 = (4-1)/2 = 3/2, gives you a circle that is not tangent to
the other two. I assume you meant (sqrt[2]-1)/2 for its radius, which
will satisfy the conditions of the problem. Then you get the following
equations:

sqrt(h^2+k^2) - (sqrt[2]-1)/2   = r
sqrt[(h-1/2)^2+(k-1/2)^2] - 1/2 = r
sqrt[(h-1/2)^2+(k+1/2)^2] - 1/2 = r

h^2 + k^2 = (r+(sqrt[2]-1)/2)^2
(h-1/2)^2 + (k-1/2)^2 = (r+1/2)^2
(h-1/2)^2 + (k+1/2)^2 = (r+1/2)^2

Subtracting the second from the third, you get k = 0, so

h^2 = (r+(sqrt[2]-1)/2)^2
h^2 - h = (r+1/2)^2 - 1/2

Subtracting the second from the first,

h = (r+(sqrt[2]-1)/2)^2 - (r+1/2)^2 + 1/2
= (2*r+sqrt[2]/2)*(sqrt[2]-2)/2 + 1/2
= (sqrt[2]-2)*(r-1/2)

Substituting this back into either of the last pair of equations gives
you a quadratic equation in r:

h^2 = (r+(sqrt[2]-1)/2)^2

so

(6-4*sqrt[2])*(r-1/2)^2 = (r+(sqrt[2]-1)/2)^2

(6-4*sqrt[2])*r^2 + (-6+4*sqrt[2])*r + (3/2-sqrt[2]) =
r^2 + (-1+sqrt[2])*r + (3/4-sqrt[2]/2)

(5-4*sqrt[2])*r^2 + (-5+3*sqrt[2])*r + (3/4-sqrt[2]/2) = 0

r^2 + ((1+5*sqrt[2])/7)*r + ((1-2*sqrt[2]/28) = 0

(r-(5-3*sqrt[2])/14)*(r-(-1-sqrt(2))/2) = 0

The second root is negative, and so extraneous. Thus

r = (5-3*sqrt[2])/14,
h = (2*sqrt[2]-1)/7.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 08/19/99 at 13:15:42
From: Doctor Rob
Subject: Re: How to find the enclosed circle

As a follow-up to my previous response, a little research reveals that
there is a very pretty formula for the radius of the enclosed circle.
If the radii of the three mutually tangent circles are r1, r2, and r3,
and the radius of the enclosed circle is r, then

1/r = 1/r1 + 1/r2 + 1/r3 + 2*sqrt[1/(r1*r2)+1/(r1*r3)+1/(r2*r3)]

1/r1 = 2/(sqrt[2]-1) = 2 + 2*sqrt[2]
1/r2 = 2
1/r3 = 2

so

1/r = 2 + 2*sqrt(2) + 2 + 2 + 2*sqrt(4+4*sqrt[2]+4+4*sqrt[2]+4)
= 6 + 2*sqrt(2) + 2*sqrt(12+8*sqrt[2])
= 6 + 2*sqrt(2) + 4*sqrt(3+2*sqrt[2])
= 6 + 2*sqrt(2) + 4*(1+sqrt[2])
= 10 + 6*sqrt(2)

r = 1/(10+6*sqrt[2])
= (10-6*sqrt[2])/(10^2-2*6^2)
= (5-3*sqrt[2])/14

which agrees with the answer I gave earlier.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Conic Sections/Circles
High School Geometry

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