JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Circle

The circle $x^2+y^2-4x-6y+11=0$ has centre $(2,3)$ and radius $\sqrt{2}$, and the tangent at $(3,2)$ has slope $1$, the line $x-y-1=0$. Rolling the circle $4$ units up this tangent shifts the centre by $4$ along the unit direction $\frac{(1,1)}{\sqrt{2}}$, placing $A=(2+2\sqrt{2},3+2\sqrt{2})$. Reflecting $A$ in the tangent gives $B=(4+2\sqrt{2},1+2\sqrt{2})$. The feet of the perpendiculars on the $x$-axis are $M=(2+2\sqrt{2},0)$ and $N=(4+2\sqrt{2},0)$, so the parallel sides are $AM=3+2\sqrt{2}$ and $BN=1+2\sqrt{2}$ with $MN=2$ between them. The trapezium area is $\frac{1}{2}(3+2\sqrt{2}+1+2\sqrt{2})\cdot 2=4(1+\sqrt{2})$.