If for a variable line \[\frac{x}{a}+\frac{y}{b}=1\], the condition \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{c}^{2}}}\] (c is a constant) is satisfied, then locus of foot of perpendicular drawn from origin to the line is [RPET 1999]

\[{{x}^{2}}-{{y}^{2}}={{c}^{2}}\]

\[{{x}^{2}}+{{y}^{2}}={{c}^{2}}/2\]

\[{{x}^{2}}+{{y}^{2}}=2{{c}^{2}}\]

\[{{x}^{2}}+{{y}^{2}}={{c}^{2}}\]

If for a variable line \[\frac{x}{a}+\frac{y}{b}=1\], the condition \[

1.	If for a variable line \[\frac{x}{a}+\frac{y}{b}=1\], the condition \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{c}^{2}}}\] (c is a constant) is satisfied, then locus of foot of perpendicular drawn from origin to the line is [RPET 1999]
A.	\[{{x}^{2}}+{{y}^{2}}={{c}^{2}}/2\]
B.	\[{{x}^{2}}+{{y}^{2}}=2{{c}^{2}}\]
C.	\[{{x}^{2}}+{{y}^{2}}={{c}^{2}}\]
D.	\[{{x}^{2}}-{{y}^{2}}={{c}^{2}}\]
Answer» D. \[{{x}^{2}}-{{y}^{2}}={{c}^{2}}\]

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If for a variable line \[\frac{x}{a}+\frac{y}{b}=1\], the condition \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{c}^{2}}}\] (c is a constant) is satisfied, then locus of foot of perpendicular drawn from origin to the line is [RPET 1999]

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