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Area of the shaded region when chord AB = 4 splits a diameter into two smaller circles - JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
In the figure alongside, a chord ABAB of a circle cuts a diameter at right angles at the point TT, and the two pieces of that diameter, of lengths 2x2x and 2y2y, are the diameters of the two smaller circles drawn inside. The chord ABAB has length 44. Find the area of the shaded region, that is, the big circle with the two smaller circles removed. Question attachment
Aπ2\dfrac{\pi}{2}
Bπ\pi
Cπ+1\pi+1
D2π2\picorrect
Solution
Step 1: ABAB and the diameter are two chords of the big circle crossing at TT, so the products of their segments are equal.
TA⋅TB=2x⋅2yTA\cdot TB = 2x \cdot 2y
 Step 2: ABAB is perpendicular to a diameter at TT, so TT bisects it and TA=TB=2TA=TB=2.
2⋅2=4xy⟹xy=12\cdot 2 = 4xy \quad\Longrightarrow\quad xy = 1
 Step 3: The big circle has diameter 2x+2y2x+2y, so radius x+yx+y; the two small circles have radii xx and yy.
Shaded=π(x+y)2−πx2−πy2=π[(x+y)2−x2−y2]=π(2xy)\text{Shaded} = \pi(x+y)^2 - \pi x^2 - \pi y^2 = \pi\left[(x+y)^2-x^2-y^2\right] = \pi\left(2xy\right)
 Step 4: Putting xy=1xy=1,
Shaded=2π(1)=2π\text{Shaded} = 2\pi(1) = 2\pi
 The answer does not depend on where TT lies on the diameter, only on AB=4AB=4. Answer: (4) 2π2\pi
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