Let a circle C₁ be obtained on rolling the circle x²+y²-4 x-6 y+11=0 upwards 4 units on the tangent T to it at the point (3,2). Let C₂ be the image of C₁ in T. Let A and B be the centers of circles C₁ and C₂ respectively, and M and N be respectively the feet of perpendiculars drawn from A and B on the x-axis. Then the area of the trapezium AMNB is :

Solution

Maths sir · 5 steps

Rolling a circle along a line just slides its centre 4 units in the tangent's direction.

1. Circle and tangent. The centre is (2, 3) and r² = 4 + 9 - 11 = 2. The tangent at (3, 2) is 3x + 2y - 2(x + 3) - 3(y + 2) + 11 = 0, that is, x - y - 1 = 0, which has slope 1.
2. Roll to get A. Moving 4 units up along slope 1 shifts the centre by 4/√2 = 2√2 in both x and y. So A = (2 + 2√2, 3 + 2√2).
3. Reflect to get B. At A, x - y - 1 = -2. The image in x - y - 1 = 0 is A - 2(-2)/2(1, -1) = A + (2, -2), so B = (4 + 2√2, 1 + 2√2).
4. Trapezium sides. AM and BN are both vertical, so they are the parallel sides: AM = 3 + 2√2 and BN = 1 + 2√2. The gap between them is MN = (4 + 2√2) - (2 + 2√2) = 2.

   Area = 1/2(AM + BN) × MN = 1/2(4 + 4√2) × 2

5. So. The area is 4 + 4√2 = 4(1 + √2), which is option (b).

The idea to keep. A circle rolling on a line keeps its centre at distance r from that line, so the centre moves parallel to the line by exactly the distance rolled. Reflecting a point in a line is a one-formula job.

Answer from the NTA final answer key. Solution by PrepWiser.