The equation of the incircle of the triangle formed by the axes and the line 4x+3y=6 is

a)x²+y²-6x-6y+9=0

b)4(x²+y²-x-y)+1=0

c)4(x²+y²+x+y)+1

d)none of these

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Please find below the solution to the asked query:


$$\begin{gathered} \mathrm{From} \mathrm{figure} \mathrm{it} \mathrm{is} \mathrm{clear} \mathrm{that} \mathrm{circle} \mathrm{will} \mathrm{lie} \mathrm{in} \mathrm{first} \mathrm{quadrant}. \\ \mathrm{Let} \mathrm{equation} \mathrm{of} \mathrm{the} \mathrm{circle} \mathrm{be} \\ {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}. \\ \mathrm{As} \mathrm{it} \mathrm{is} \mathrm{incircle} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{formed} \mathrm{by} \mathrm{the} \mathrm{axes} \mathrm{and} \mathrm{the} \mathrm{line} 4\mathrm{x}+3\mathrm{y}=6. \\ \mathrm{Hence} \mathrm{it} \mathrm{touches} \mathrm{x}-\mathrm{axis}, \mathrm{y}-\mathrm{axis} \mathrm{and} 4\mathrm{x}+3\mathrm{y}=6 \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ \mathrm{When} \mathrm{circle} \mathrm{touches} \mathrm{x}-\mathrm{axis} \mathrm{then}, \\ \mathrm{r}=\left|\mathrm{f}\right| ,\mathrm{where} \mathrm{r} \mathrm{is} \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{circle}. \\ \mathrm{When} \mathrm{circle} \mathrm{touches} \mathrm{y}-\mathrm{axis} \mathrm{then}, \\ \mathrm{r}=\left|\mathrm{g}\right| \\ \Rightarrow \left|\mathrm{g}\right|=\left|\mathrm{f}\right| \\ \Rightarrow \mathrm{g}=\mathrm{f} \\ \mathrm{Also} \\ \mathrm{r}=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}} \\ \mathrm{As} \mathrm{r}=\left|\mathrm{g}\right| \\ \left|\mathrm{g}\right|=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}} \\ \Rightarrow {\mathrm{g}}^2={\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c} \\ \Rightarrow \mathrm{c}={\mathrm{f}}^2 \\ \mathrm{Now} \mathrm{as} 4\mathrm{x}+3\mathrm{y}-6=0 \mathrm{also} \mathrm{touches} \mathrm{the} \mathrm{circle}, \mathrm{hence} \mathrm{perpendicular} \mathrm{from} \mathrm{centre} \mathrm{of} \mathrm{the} \mathrm{circle} \\ \mathrm{to} \mathrm{it} \mathrm{must} \mathrm{be} \mathrm{equal} \mathrm{to} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{circle}. \\ \mathrm{Centre} \mathrm{of} \mathrm{circle} \mathrm{is} \left(-\mathrm{g},-\mathrm{f}\right) \\ \Rightarrow \mathrm{r}=\left|\frac{-4\mathrm{g}-3\mathrm{f}-6}{\sqrt{4^2+3^2}}\right| \\ \Rightarrow \mathrm{r}=\left|\frac{-4\mathrm{g}-3\mathrm{g}-6}{\sqrt{25}}\right| \\ \Rightarrow \mathrm{r}=\left|\frac{-7\mathrm{g}-6}{5}\right| \\ \Rightarrow \mathrm{r}=\left|\frac{7\mathrm{g}+6}{5}\right| \\ \Rightarrow \left|\mathrm{g}\right|=\left|\frac{7\mathrm{g}+6}{5}\right| \\ \Rightarrow \mathrm{g}=\pm \left(\frac{7\mathrm{g}+6}{5}\right) \\ \Rightarrow 5\mathrm{g}=\pm \left(7\mathrm{g}+6\right) \\ \mathrm{As} \mathrm{circle} \mathrm{lies} \mathrm{in} \mathrm{first} \mathrm{quadrant}, \mathrm{hence} -\mathrm{g} \mathrm{will} \mathrm{be} \mathrm{positive} \mathrm{and} \mathrm{g} \mathrm{will} \mathrm{be} \mathrm{negative} \\ \mathrm{Hence} \mathrm{considering} \mathrm{positive} \mathrm{sign} \mathrm{we} \mathrm{get}, \\ \Rightarrow 5\mathrm{g}=7\mathrm{g}+6 \\ \Rightarrow -2\mathrm{g}=6 \\ \Rightarrow \mathrm{g}=-3 \\ \Rightarrow \mathrm{f}=-3 \\ \Rightarrow \mathrm{r}=3 \\ {\mathrm{g}}^2=\mathrm{c} \\ \Rightarrow \mathrm{c}=9 \\ \mathbf{Hence}\mathbf{ }\mathbf{equation}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{circle}\mathbf{ }\mathbf{is} \\ {\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{6}\mathbf{x}\mathbf{-}\mathbf{6}\mathbf{xy}\mathbf{+}\mathbf{9}\mathbf{=}\mathbf{0} \\ \mathbf{Hence}\mathbf{ }\mathbf{option}\left(\mathbf{i}\right)\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{correct}\mathbf{.} \end{gathered}$$

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