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.

Problem : 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.

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 - 2sqrt(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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