Question

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 AMNB is:

• A. $2(2+\sqrt{2})$
• B. $2(1+\sqrt{2})$
• C. $4(1+\sqrt{2})$
• D. $3+2 \sqrt{2}$

Answer 1

The Correct Option is C

The correct answer is (C) : \(4(1+\sqrt2)\)

\(C=(2,3),r=\sqrt2​\) 
Centre of G=A\(=2+\frac{\sqrt4}2​​, \)
\(3+\frac{\sqrt4}{2}​=(2+2\sqrt2​,3+2\sqrt2​) \)
\(A(2+2\sqrt2​,3+2\sqrt2​) \)
\(B(4+2\sqrt2​,1+2\sqrt2​) \)
\(\frac{x−(2+2\sqrt2​)​}{1}=\frac{y−(3+2\sqrt2​)​}{-1}=2 \)
∴ area of trapezium: 
\(=\frac{1}{2}​(4+4\sqrt2​)2\)
\(=4(1+\sqrt2)\)