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.
In the figure below:
Find the radius of the larger circle.
We present two different solution methods.
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:
\[ AN = NT \] \[ 2AN^2 = 16 \] \[ AN^2 = 8 \] \[ AN = NT = 2\sqrt{2} \]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:
\[ (r - 2\sqrt{2})^2 + (r - 2\sqrt{2})^2 = r^2 \] \[ 2(r - 2\sqrt{2})^2 = r^2 \]Taking the square root:
\[ \sqrt{2}(r - 2\sqrt{2}) = r \]Solving gives:
\[ r = 4 + 4\sqrt{2} \]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 yields two solutions, but only the positive value is valid:
\[ r = 4 + 4\sqrt{2} \]Explore more tutorials and interactive geometry problems: