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
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
AN^{2} + NT^{2} = 4^{2}

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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