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

Question

Accepted Answer

${{x}^{2}}-{{y}^{2}}={{c}^{2}}$

Answer

${{x}^{2}}+{{y}^{2}}={{c}^{2}}/2$

Answer

${{x}^{2}}+{{y}^{2}}=2{{c}^{2}}$

Answer

${{x}^{2}}+{{y}^{2}}={{c}^{2}}$

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Question: 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

st-line

Question: 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

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