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, an

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Solution

Q.
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: [2023]

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

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

3 $4 (1 + \sqrt{2})$

4 $3 + 2 \sqrt{2}$

Ans.
(3)

We have, $x^{2} + y^{2} - 4 x - 6 y + 11 = 0$ ...(i)

Centre = (2, 3); Radius $= \sqrt{2}$

Tangent at $(3, 2)$ is, $3 x + 2 y - 2 (x + 3) - 3 (y + 2) + 11 = 0$

$\Rightarrow x - y = 1$

On rolling the given circle (i) upwards 4 units on the tangent $x - y - 1 = 0$ , centre of the circle also moves upwards 4 units on $T$ . Let the centre of the new circle $C_{1}$ be $(h, k)$ .

Since, slope of tangent = Slope of line joining two centres of the circle.

Then the increment in both coordinates will be same.

From figure,

${(2 + a - 2)}^{2} + {(3 + a - 3)}^{2} = 16$

$a^{2} + a^{2} = 16 \Rightarrow a^{2} = 8 \Rightarrow a = 2 \sqrt{2}$

Hence, the centre of circle $C_{1}$ is $(2 + 2 \sqrt{2}, 3 + 2 \sqrt{2})$ and radius remains same $\sqrt{2}$ .

Equation of circle $C_{1}$ is ${(x - 2 - 2 \sqrt{2})}^{2} + {(y - 3 - 2 \sqrt{2})}^{2} = 2$

The centre of circle $C_{2}$ is the image of $C_{1}$ in $T$ , then

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

$\Rightarrow x - 2 - 2 \sqrt{2} = 2 and y - 3 - 2 \sqrt{2} = - 2$

$\Rightarrow y = 1 + 2 \sqrt{2} \Rightarrow x = 4 + 2 \sqrt{2}$

The centre of circle $C_{2}$ is $B (4 + 2 \sqrt{2}, 1 + 2 \sqrt{2}) .$

Feet of perpendicular from $A$ and $B$ on the $x$ -axis are $M (2 + 2 \sqrt{2}, 0) and N (4 + 2 \sqrt{2}, 0)$ respectively.

Area of trapezium AMNB $= \frac{1}{2} (3 + 2 \sqrt{2} + 1 + 2 \sqrt{2}) (4 + 2 \sqrt{2} - 2 - 2 \sqrt{2})$

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

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