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 :
Solution
Given circle is $x^{2}+y^{2}-4 x-6 y+11=0$, centre $(2,3)$
<br/><br/>Tangent at $(3,2)$ is $x-y=1$
<br/><br/>After rolling up by 4 units centre of $C_{1}$ is
<br/><br/>$A \equiv\left(2+\frac{4}{\sqrt{2}}, 3+\frac{4}{\sqrt{2}}\right)$
<br/><br/>$\Rightarrow A=(2+2 \sqrt{2}, 3+2 \sqrt{2})$
<br/><br/>$B$ is the image of $A$ in $x-y=1$
<br/><br/>$\frac{x-(2+2 \sqrt{2})}{1}=\frac{y-(3+2 \sqrt{2})}{-1}=\frac{-2(-2)}{2}=2$
<br/><br/>$\Rightarrow x=4+2 \sqrt{2}, y=1+2 \sqrt{2}$
<br/><br/>Area of $A M N B$
<br/><br/>$=\frac{1}{2}(4+4 \sqrt{2})(4+2 \sqrt{2}-(2+2 \sqrt{2}))$
<br/><br/>$=4(1+\sqrt{2})$
About this question
Subject: Mathematics · Chapter: Circles · Topic: Equation of a Circle
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