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…

Question

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 $$\mathrm{T}$$ to it at the point $(3,2)$. Let $C_{2}$ be the image of $C_{1}$ in $$\mathrm{T}$$. 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 :

Accepted Answer

Given circle is $x^{2}+y^{2}-4 x-6 y+11=0$, centre $(2,3)$

Tangent at $(3,2)$ is $x-y=1$

After rolling up by 4 units centre of $C_{1}$ is

$A \equiv\left(2+\frac{4}{\sqrt{2}}, 3+\frac{4}{\sqrt{2}}\right)$

$\Rightarrow A=(2+2 \sqrt{2}, 3+2 \sqrt{2})$

$B$ is the image of $A$ in $x-y=1$

$\frac{x-(2+2 \sqrt{2})}{1}=\frac{y-(3+2 \sqrt{2})}{-1}=\frac{-2(-2)}{2}=2$

$\Rightarrow x=4+2 \sqrt{2}, y=1+2 \sqrt{2}$

Area of $A M N B$

$=\frac{1}{2}(4+4 \sqrt{2})(4+2 \sqrt{2}-(2+2 \sqrt{2}))$

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

Solutions · 1