How do you find the radius of a larger circle tangent to a smaller circle inside a square?

Given the small circle's perimeter of 4π, its diameter is 4. By applying the Pythagorean theorem using either a constructed square for the small circle or a larger right triangle connecting the centers and vertices, we set up quadratic equations that solve to r = 4 + 4√2.

Two Tangent Circles Inside a Square – Geometry Problem & Solution

You are given the perimeter of a small circle and asked to determine the radius of a larger circle inscribed in a square. The two circles are tangent to each other.

Problem Statement

In the figure below:

The small circle with center B and the larger circle with center C are tangent at point T.
Point A is a vertex of the square circumscribing the larger circle.
Points A, B, T, and C are collinear.
The perimeter of the small circle is \(4\pi\).

Find the radius of the larger circle.

Two tangent circles inside a square

Solution

We present two different solution methods.

Method 1: Using a Constructed Square and Pythagorean Theorem

From point \(T\), draw segments to the intersection points \(M\) and \(N\) of the small circle with the square. Because \(AT\) is a diameter, angles \(AMT\) and \(ANT\) are right angles. By symmetry, \(AM = AN\), so \(AMTN\) is a square.

The diagonal of this smaller square equals the diameter of the small circle.

Since the perimeter of the small circle is \(4\pi\):

\[ d = \frac{\text{perimeter}}{\pi} = \frac{4\pi}{\pi} = 4 \]

Thus the diagonal of the small square is 4. Using the Pythagorean theorem in triangle \(ANT\):

\[ AN^2 + NT^2 = 4^2 \]

Because the square has equal sides (\(AN = NT\)):

\[ 2AN^2 = 16 \implies AN^2 = 8 \implies AN = NT = 2\sqrt{2} \]

Now consider right triangle \(TPC\). Using the Pythagorean theorem:

\[ x^2 + y^2 = r^2 \]

Observe that:

\[ x = r - 2\sqrt{2}, \quad y = r - 2\sqrt{2} \]

Substitute into the Pythagorean relation:

\[ (r - 2\sqrt{2})^2 + (r - 2\sqrt{2})^2 = r^2 \implies 2(r - 2\sqrt{2})^2 = r^2 \]

Taking the square root:

\[ \sqrt{2}(r - 2\sqrt{2}) = r \implies \sqrt{2}r - 4 = r \implies r(\sqrt{2} - 1) = 4 \]

Solving gives:

\[ r = \frac{4}{\sqrt{2} - 1} = 4(\sqrt{2} + 1) = 4 + 4\sqrt{2} \]

Method 2: Direct Use of a Larger Right Triangle

Consider right triangle \(AQC\). Applying the Pythagorean theorem:

\[ r^2 + r^2 = (r + 4)^2 \] \[ 2r^2 = r^2 + 8r + 16 \] \[ r^2 - 8r - 16 = 0 \]

Solving the quadratic equation using the quadratic formula yields:

\[ r = \frac{8 \pm \sqrt{(-8)^2 - 4(1)(-16)}}{2} = \frac{8 \pm \sqrt{64 + 64}}{2} = \frac{8 \pm \sqrt{128}}{2} = 4 \pm 4\sqrt{2} \]

Only the positive value is valid:

\[ r = 4 + 4\sqrt{2} \]

Final Answer

\[ \boxed{r = 4 + 4\sqrt{2}} \]

More Geometry Resources

Explore more tutorials and interactive geometry problems:

Geometry Tutorials, Problems and Interactive Applets