Let a circle $C_{1}$ be obtained on rolling the circle $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$ upwards 4 units on the tangent $T$ to it at the point (3, 2). Let $C_{2}$ be the image of $C_{1}$ in $T$ . Let $A$ and $B$ be the centers of circles $C_{1}$ and $C_{2}$ 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 $A M N B$ is:

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$2 (1 + \sqrt{2})$

$4 (1 + \sqrt{2})$

$3 + 2 \sqrt{2}$

$2 (2 + \sqrt{2})$

answer is D.

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Detailed Solution

Equation of tangent at (3, 2) is x − y −1= 0
Slope of $\vec{C A} is 1, CA = 4$ units coordinates of coordinates of
$A = (2 + 4 (\frac{1}{\sqrt{2}}), 3 + 4 (\frac{1}{\sqrt{2}})) = (2 + 2 \sqrt{2}, 3 + 2 \sqrt{2})$
Coordinates of $B = (2 + 2 \sqrt{2} - (2 \sqrt{2}) (\frac{- 1}{\sqrt{2}}), 3 + 2 \sqrt{2} - (2 \sqrt{2}) (\frac{1}{\sqrt{2}}))$
$= (4 + 2 \sqrt{2}, 1 + 2 \sqrt{2})$
$∴$ Area of trapezium AMNB $= \frac{1}{2} (3 + 2 \sqrt{2} + 1 + 2 \sqrt{2}) (2)$
$= 4 (\sqrt{2} + 1)$

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