### Circle Question 8

#### Question 8 - 31 January - Shift 1

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 :

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

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

(3) $3+2 \sqrt{2}$

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

## Show Answer

#### Answer: (2)

#### Solution:

#### Formula: Foot of the perpendicular from a point on the line, Equation of a straight line in two points form

$C=(2,3), r=\sqrt{2}$

Centre of $G=A=2+ \frac{4}{\sqrt{2}}$,

$3+\frac{4}{\sqrt{2}}=(2+2 \sqrt{2}, 3+2 \sqrt{2})$

$A(2+2 \sqrt{2}, 3+2 \sqrt{2})$

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

$\frac{x-(2+2 \sqrt{2})}{1}=\frac{y-(3+2 \sqrt{2})}{-1}=2$

$\therefore$ area of trapezium:

$\frac{1}{2}\times (4+4 \sqrt{2}) \times 2=4(1+\sqrt{2})$