Question

Let a circle $C_1$ be obtained on rolling the circle $x^2+y^2-4x-6y+11=0$ 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

Answer: (C)

$C=(2,3),r=\sqrt{2}$
Centre of $G=A=2+4\frac{1}{\sqrt{2}},$
$3+\frac{4}{\sqrt{2}}=(2+2\sqrt{2},3+2\sqrt{2})$
$A(2+2\sqrt{2},3+2\sqrt{2})$
$B(4+2\sqrt{2},1+2\sqrt{2})$
$\frac{x-(2+2\sqrt{2})}{1}=\frac{y-(3+2\sqrt{2})}{-1}=2$
$\therefore \text{ area of trapezium: }$
$\frac{1}{2}(4+4\sqrt{2})2=4(1+\sqrt{2})$