Let a circle C_1 be obtained by rolling the c | vedclass

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(B) The given circle is $x^2+y^2-4x-6y+11=0$. Its center is $C(2,3)$ and radius $r = \sqrt{2^2+3^2-11} = \sqrt{4+9-11} = \sqrt{2}$.The tangent $T$ at

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

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(B) The given circle is $x^2+y^2-4x-6y+11=0$. Its center is $C(2,3)$ and radius $r = \sqrt{2^2+3^2-11} = \sqrt{4+9-11} = \sqrt{2}$.The tangent $T$ at $(3,2)$ to the circle $(x-2)^2+(y-3)^2=2$ is $(3-2)(x-2)+(2-3)(y-3)=2$,which simplifies to $1(x-2)-1(y-3)=2$,or $x-y-1=0$.The slope of the tangent is $m=1$. The unit vector along the tangent is $\frac{1}{\sqrt{2}}(1,1)$.Rolling the circle $4$ units along the tangent means moving the center $C(2,3)$ by a vector $4 	imes \frac{1}{\sqrt{2}}(1,1) = (2\sqrt{2}, 2\sqrt{2})$.Thus,the center $A$ of $C_1$ is $(2+2\sqrt{2}, 3+2\sqrt{2})$.$C_2$ is the image of $C_1$ in the line $x-y-1=0$. The center $B$ is the reflection of $A$ in $x-y-1=0$.Using the reflection formula $\frac{x_B-x_A}{1} = \frac{y_B-y_A}{-1} = -2 \frac{x_A-y_A-1}{1^2+(-1)^2} = -(x_A-y_A-1)$.$x_A-y_A-1 = (2+2\sqrt{2})-(3+2\sqrt{2})-1 = -2$.So,$\frac{x_B-(2+2\sqrt{2})}{1} = \frac{y_B-(3+2\sqrt{2})}{-1} = -(-2) = 2$.$x_B = 2+2\sqrt{2}+2 = 4+2\sqrt{2}$ and $y_B = 3+2\sqrt{2}-2 = 1+2\sqrt{2}$.$A = (2+2\sqrt{2}, 3+2\sqrt{2})$ and $B = (4+2\sqrt{2}, 1+2\sqrt{2})$.The coordinates of $M$ and $N$ are $(2+2\sqrt{2}, 0)$ and $(4+2\sqrt{2}, 0)$.The area of the trapezium $AMNB$ is $\frac{1}{2} 	imes (AM+BN) 	imes MN$.$AM = 3+2\sqrt{2}$,$BN = 1+2\sqrt{2}$,$MN = (4+2\sqrt{2})-(2+2\sqrt{2}) = 2$.Area $= \frac{1}{2} 	imes (3+2\sqrt{2}+1+2\sqrt{2}) 	imes 2 = 4+4\sqrt{2} = 4(1+\sqrt{2})$.

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