If the lines represented by the equation ax^2 + 2hxy + by^2 + 2gx +2fy + c=0 are equidistant from origin , then
1.f^4 – g^4 = c (af^2 – bg^2)
2. f^4 – g^4 = c (bf^2 – ag^2)
3. f^4 – g^4 = -c (bf^2 – ag^2)
4.none

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0 \\ \mathrm{Let} \mathrm{lines} \mathrm{by} \mathrm{y}-{\mathrm{m}}_1\mathrm{x}-{\mathrm{c}}_1=0 \mathrm{and} \mathrm{y}-{\mathrm{m}}_2\mathrm{x}-{\mathrm{c}}_2=0 \\ \mathrm{Both} \mathrm{lines} \mathrm{are} \mathrm{equidistant} \mathrm{from} \mathrm{origin}, \mathrm{hence} \\ \frac{\left|{\mathrm{c}}_1\right|}{\sqrt{1+{{\mathrm{m}}_1}^2}}=\frac{\left|{\mathrm{c}}_2\right|}{\sqrt{1+{{\mathrm{m}}_2}^2}} \\ \Rightarrow \frac{{{\mathrm{c}}_1}^2}{1+{{\mathrm{m}}_1}^2}=\frac{{{\mathrm{c}}_2}^2}{1+{{\mathrm{m}}_2}^2} \\ \Rightarrow {{\mathrm{c}}_1}^2+{{\mathrm{c}}_1}^2{{\mathrm{m}}_2}^2={{\mathrm{c}}_2}^2+{{\mathrm{c}}_2}^2{{\mathrm{m}}_1}^2 \\ \Rightarrow {{\mathrm{c}}_2}^2{{\mathrm{m}}_1}^2-{{\mathrm{c}}_1}^2{{\mathrm{m}}_2}^2={{\mathrm{c}}_1}^2-{{\mathrm{c}}_2}^2 \\ \Rightarrow \left({\mathrm{m}}_1{\mathrm{c}}_2+{\mathrm{m}}_2{\mathrm{c}}_1\right)\left({\mathrm{m}}_1{\mathrm{c}}_2-{\mathrm{m}}_2{\mathrm{c}}_1\right)=\left({\mathrm{c}}_1+{\mathrm{c}}_2\right)\left({\mathrm{c}}_1-{\mathrm{c}}_2\right) \\ \Rightarrow \left({\mathrm{m}}_1{\mathrm{c}}_2+{\mathrm{m}}_2{\mathrm{c}}_1\right)\sqrt{{\left({\mathrm{m}}_1{\mathrm{c}}_2+{\mathrm{m}}_2{\mathrm{c}}_1\right)}^2-4{\mathrm{m}}_1{\mathrm{m}}_2{\mathrm{c}}_1{\mathrm{c}}_2}=\left({\mathrm{c}}_1+{\mathrm{c}}_2\right)\sqrt{{\left({\mathrm{c}}_1+{\mathrm{c}}_2\right)}^2-4{\mathrm{c}}_1{\mathrm{c}}_2}...\left(\mathrm{i}\right) \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ {\mathrm{m}}_1{\mathrm{c}}_2+{\mathrm{m}}_2{\mathrm{c}}_1=\frac{2\mathrm{g}}{\mathrm{b}} \\ {\mathrm{c}}_1+{\mathrm{c}}_2=-\frac{2\mathrm{f}}{\mathrm{b}} \\ {\mathrm{c}}_1{\mathrm{c}}_2=\frac{\mathrm{c}}{\mathrm{b}} \\ {\mathrm{m}}_1{\mathrm{m}}_2=\frac{\mathrm{a}}{\mathrm{b}} \\ \mathrm{Put} \mathrm{in} \left(\mathrm{i}\right) \\ \Rightarrow \left(\frac{2\mathrm{g}}{\mathrm{b}}\right)\sqrt{\frac{4{\mathrm{g}}^2}{{\mathrm{b}}^2}-\frac{4\mathrm{ac}}{{\mathrm{b}}^2}}=\left(-\frac{2\mathrm{f}}{\mathrm{b}}\right)\sqrt{\frac{4{\mathrm{f}}^2}{{\mathrm{b}}^2}-\frac{4\mathrm{c}}{\mathrm{b}}} \\ \Rightarrow \mathrm{g}\sqrt{\frac{{\mathrm{g}}^2}{{\mathrm{b}}^2}-\frac{\mathrm{ac}}{{\mathrm{b}}^2}}=-\mathrm{f}\sqrt{\frac{{\mathrm{f}}^2}{{\mathrm{b}}^2}-\frac{\mathrm{bc}}{{\mathrm{b}}^2}} \\ \mathrm{Square} \mathrm{both} \mathrm{sides} \\ \Rightarrow {\mathrm{g}}^2\left(\frac{{\mathrm{g}}^2-\mathrm{ac}}{{\mathrm{b}}^2}\right)={\mathrm{f}}^2\left(\frac{{\mathrm{f}}^2-\mathrm{bc}}{{\mathrm{b}}^2}\right) \\ \Rightarrow {\mathrm{g}}^4-{\mathrm{acg}}^2={\mathrm{f}}^4-{\mathrm{bcf}}^2 \\ \mathbf{\Rightarrow }{\mathbf{f}}^{\mathbf{4}}\mathbf{-}{\mathbf{g}}^{\mathbf{4}}\mathbf{=}\mathbf{c}\left({\mathbf{bf}}^{\mathbf{2}}\mathbf{-}{\mathbf{ag}}^{\mathbf{2}}\right)\mathbf{ }\left[\mathbf{Answer}\right] \end{gathered}$$

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