The circle $x^2+y^2-4 x-8 y+16=0$ rolls up along the tangent drawn to it at $(2+\sqrt{3}, 3)$ by 2 units. The equation of the circle in the new position is

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Aicharya · Accepted Answer

$x^2+y^2-4 x-8 y+16=0$

$C \equiv(2,4)$ $\begin{aligned} & m_{\mathrm{CP}}=\frac{4-3}{2-2-\sqrt{3}}=-\frac{1}{\sqrt{3}} \\ & m_{\mathrm{CP}} \times m_{\mathrm{CC}^{\prime}}=-1\end{aligned}$ $\begin{aligned} \Rightarrow & m_{\mathrm{CC}}=\sqrt{3} \\ \Rightarrow & \tan \theta=\sqrt{3}=\tan 60^{\circ} \\ \Rightarrow & \theta=60^{\circ} \\ & \mathrm{CC}^{\prime}=2\end{aligned}$ Then, $C^{\prime} \equiv\left(2+2 \cos 60^{\circ}, 4+2 \sin 60^{\circ}\right)$ $\therefore \quad C^{\prime} \equiv(3,4+\sqrt{3})$ $C^{\prime} P^{\prime}=2$ Equation of new circle : $\begin{aligned} & (x-3)^2+(y-4-\sqrt{3})^2=2^2 \\ & \Rightarrow x^2+y^2-6 x-2(4+\sqrt{3}) y+9+(4+\sqrt{3})^2=4 \\ & \Rightarrow x^2+y^2-6 x-2(4+\sqrt{3}) y+(24+8 \sqrt{3})=0\end{aligned}$

The circle $x^2+y^2-4 x-8 y+16=0$ rolls up along the tangent drawn to it at $(2+\sqrt{3}, 3)$ by 2 units.…

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