JEE Main 2023 (Online) 31st January Morning ShiftQuestion 1
Let a circle $C_1$ be obtained on rolling the circle $x^2+y^2-4x-6y+11=0$ 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:
Circle
Solution
The circle has centre $(2,3)$, radius $\sqrt2$; the tangent at $(3,2)$ is $x-y=1$ (direction $(1,1)$). Rolling up $4$ units moves the centre to $A=\left(2+2\sqrt2,\,3+2\sqrt2\right)$. Its image in $x-y=1$ is $B=\left(4+2\sqrt2,\,1+2\sqrt2\right)$. Then $AM=3+2\sqrt2$, $BN=1+2\sqrt2$, $MN=2$, and the area is $\frac12(4+4\sqrt2)\cdot2=4\left(1+\sqrt2\right)$. (The source lost the brackets in the options; restored as in the original paper.)