Two Tangent Circles and a Square- Problem With Solution

You are givn the perimeter of a small circle to find the radius of a larger circle inscribed within a square.


In the figure below, the small circle with center B and the larger circle with center C are tangent at point T. A is the vertex of the square circumscribing the larger circle. Points A, B, T and C are collinear. The perimeter of the small circle is equal to 4π. Find the radius of the larger circle.
two tangent circles and a square

Solution to Problem :

    2 Methods to solve the above problem
    METHOD 1
  • From point T, we draw TM and TN where points M and N are points of intersection of the small circle with the large square. Since AT is a diameter, AMT and ANT are right angles. Because of symmetry, the lengths of the cords AM and AN are equal and therefore AMTN is a square. The diagonal of this small square is equal to the diameter of the small circle. Since we know the perimeter, the diameter d is given by
    two tangent circles and a square, solution

    d = perimeter / π = 4 π / π = 4
  • Now that we have the diagonal of the small square, we find the length of its sides as follows
    AN2 + NT2 = 42
  • Solve to obtain
    AN = NT = 2sqrt(2)
  • We now consider the rigth triangle TPC and use Pythagora's theorem to write
    x 2 + y 2 = r 2
  • Note that
    x = r - AN = r - 2sqrt(2) and y = r - NT = r - 2sqrt(2)
  • Substitute x and y in the equation x 2 + y 2 = r 2 and write
    (r - 2sqrt(2)) 2 + (r - 2sqrt(2)) 2 = r 2
  • Group like term
    2 (r - 2 sqrt(2)) 2 = r 2
  • Extract the square root to obtain two equations
    sqrt(2) (r - 2sqrt(2)) = r or sqrt(2) (r - 2sqrt(2)) = - r
  • The above equations gives two solutions but only one is valid and is given by.
    r = 4 + 4sqrt(2)
    METHOD 2
  • Consider the right triangle AQC and use Pythagora's theorem to write
    r 2 + r 2 = (r + 4) 2
  • Expand and solve to obtain 2 solutions to the above equation but again only one of them is valid and is given by r = 4 + 4sqrt(2).

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