Let a circle C_1 be obtained by rolling the circle x^2+y^2-4x-6y+11=0 upwards 4 units alon…

Question

Let a circle $C_1$ be obtained by rolling the circle $x^2+y^2-4x-6y+11=0$ upwards $4$ units along 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 centres of circles $C_1$ and $C_2$ respectively, and $M,N$ be the feet of perpendiculars from $A,B$ onto the $x$-axis. Then the area of trapezium $AMNB$ is:

Accepted Answer

Original circle: centre $(2,3)$, radius $\sqrt2$. Radius to $(3,2)$ is along $(1,-1)$, so the tangent $T$ at $(3,2)$ has direction $(1,1)/\sqrt2$ and equation $x-y=1$.

Rolling the centre $4$ units along $T$: $A=(2,3)+4\cdot\dfrac{(1,1)}{\sqrt2}=(2+2\sqrt2,\,3+2\sqrt2)$.

$B$ is the reflection of $A$ in $T:x-y-1=0$. With $d=\dfrac{x_0-y_0-1}{2}=\dfrac{-1-1}{2}=-1$:
$$B=(x_0+2,\ y_0-2)=(4+2\sqrt2,\ 1+2\sqrt2).$$

$M=(2+2\sqrt2,0)$, $N=(4+2\sqrt2,0)$; $AM=3+2\sqrt2$, $BN=1+2\sqrt2$, $MN=2$.

$$\text{Area}=\frac12(AM+BN)\cdot MN=\frac12(4+4\sqrt2)(2)=4(1+\sqrt2).$$

Solutions · 1