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:
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:
Show Hint
- $2(2+\sqrt{2})$
- $2(1+\sqrt{2})$
- $4(1+\sqrt{2})$
- $3+2 \sqrt{2}$
The Correct Option is C
Approach Solution - 1
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:
\(=\frac{1}{2}(4+4\sqrt2)2\)
\(=4(1+\sqrt2)\)
Approach Solution -2
The equation of the circle \( C_1 \) is: \[ x^2 + y^2 - 4x - 6y + 11 = 0. \] Rewriting in standard form, we complete the square: \[ (x - 2)^2 + (y - 3)^2 = 4. \] Thus, the center of \( C_1 \) is \( (2, 3) \), and the radius is 2. The new circle \( C_2 \) is obtained by rolling the circle upwards 4 units along the tangent to \( C_1 \). Hence, the center of \( C_2 \) will be at \( (3, 2) \), and the radius remains the same. Now, the centers of the two circles are \( A(2 + 2\sqrt{2}, 3 + 2\sqrt{2}) \) and \( B(4 + 2\sqrt{2}, 1 + 2\sqrt{2}) \). The area of trapezium \( AMNB \) can be calculated as: \[ \text{Area of trapezium} = \frac{1}{2} \left( 4 + 4\sqrt{2} \right) = 4 \left( 1 + \sqrt{2} \right). \]
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Concepts Used:
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Diagram Explaining Tangents and Normal:
