if (alpha,beta) is a point on the circle whose centre is on the x-axis and which touches the line x+y=0 at (2,-2) ,then find the greatest integral value of alpha

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$$\begin{gathered} \mathrm{Equation} \mathrm{of} \mathrm{circle} \mathrm{is} {\left(\mathrm{x}-\mathrm{h}\right)}^2+{\left(\mathrm{y}-\mathrm{k}\right)}^2={\mathrm{r}}^2 \\ \mathrm{As} \mathrm{centre} \mathrm{of} \mathrm{circle} \mathrm{lies} \mathrm{on} \mathrm{X} \mathrm{axis}, \mathrm{hence} \mathrm{k}=0 \\ \Rightarrow {\left(\mathrm{x}-\mathrm{h}\right)}^2+{\mathrm{y}}^2={\mathrm{r}}^2 ; \mathrm{equation}\left(\mathrm{i}\right) \\ \mathrm{Now} \\ \mathrm{Slope} \mathrm{of} \mathrm{line} \mathrm{x}+\mathrm{y}=2 \mathrm{is} \\ \mathrm{m}=-\frac{\mathrm{Coefficient} \mathrm{of} \mathrm{x}}{\mathrm{Coefficient} \mathrm{of} \mathrm{y}}=-1={\frac{\mathrm{dy}}{\mathrm{dx}}}_{\left(2,-2\right)} \\ \mathrm{Differentiating} \left(\mathrm{i}\right), \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \mathrm{we} \mathrm{get}, \\ 2\left(\mathrm{x}-\mathrm{h}\right)+2\mathrm{y}.\frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2\left(\mathrm{x}-\mathrm{h}\right)}{2\mathrm{y}}=\frac{-\left(\mathrm{x}-\mathrm{h}\right)}{\mathrm{y}} \\ \Rightarrow {\frac{\mathrm{dy}}{\mathrm{dx}}}_{\left(2,-2\right)}=\frac{-\left(2-\mathrm{h}\right)}{-2}=\frac{2-\mathrm{h}}{2} \\ \Rightarrow \frac{2-\mathrm{h}}{2}=-1 \\ \Rightarrow 2-\mathrm{h}=-2 \\ \Rightarrow \mathrm{h}=4 \\ \Rightarrow {\left(\mathrm{x}-4\right)}^2+{\mathrm{y}}^2={\mathrm{r}}^2 \\ \mathrm{Now} \left(2.-2\right) \mathrm{lies} \mathrm{on} \mathrm{circle}. \\ \Rightarrow {\left(2-4\right)}^2+{\left(-2\right)}^2={\mathrm{r}}^2 \\ \Rightarrow 4+4={\mathrm{r}}^2 \\ \Rightarrow {\mathrm{r}}^2=8 \\ \Rightarrow \mathrm{r}=2\sqrt{2} \\ \Rightarrow {\left(\mathrm{x}-4\right)}^2+{\mathrm{y}}^2={\left(2\sqrt{2}\right)}^2 \\ \mathrm{Any} \mathrm{general} \mathrm{point} \mathrm{on} \mathrm{this} \mathrm{circle} \mathrm{will} \mathrm{be} \left(4+2\sqrt{2}\mathrm{cos\theta },2\sqrt{2}\mathrm{sin\theta }\right) \\ \Rightarrow \left(\mathrm{\alpha },\mathrm{\beta }\right)=\left(4+2\sqrt{2}\mathrm{cos\theta },2\sqrt{2}\mathrm{sin\theta }\right) \\ \Rightarrow \mathrm{\alpha }=4+2\sqrt{2}\mathrm{cos\theta } \\ \mathrm{Now} \mathrm{maximum} \mathrm{value} \mathrm{of} \mathrm{cos\theta }=1 \\ \Rightarrow {\mathrm{\alpha }}_{\max }=4+2\sqrt{2}=6.828 \\ \mathbf{Hence}\mathbf{ }\mathbf{greatest}\mathbf{ }\mathbf{integral}\mathbf{ }\mathbf{value}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{\alpha }\mathbf{ }\mathbf{will}\mathbf{ }\mathbf{be}\mathbf{ }\mathbf{6}\mathbf{.} \end{gathered}$$

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