If for a variable line frac{x}{a} + frac{y}{b | vedclass

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(C) The equation of the given line is $\frac{x}{a} + \frac{y}{b} = 1$ .....$(i)$The slope of this line is $m_1 = -\frac{b}{a}$.The slope of the perpen

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

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(C) The equation of the given line is $\frac{x}{a} + \frac{y}{b} = 1$ .....$(i)$The slope of this line is $m_1 = -\frac{b}{a}$.The slope of the perpendicular line passing through the origin $(0, 0)$ is $m_2 = \frac{a}{b}$.The equation of the perpendicular line is $y - 0 = \frac{a}{b}(x - 0)$,which simplifies to $\frac{x}{b} - \frac{y}{a} = 0$ .....$(ii)$Let $(x, y)$ be the foot of the perpendicular. Squaring and adding equations $(i)$ and $(ii)$:$\left(\frac{x}{a} + \frac{y}{b}\right)^2 + \left(\frac{x}{b} - \frac{y}{a}\right)^2 = 1^2 + 0^2$$x^2\left(\frac{1}{a^2} + \frac{1}{b^2}\right) + y^2\left(\frac{1}{b^2} + \frac{1}{a^2}\right) = 1$Since $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$,we substitute this into the equation:$(x^2 + y^2)\left(\frac{1}{c^2}\right) = 1$Therefore,the locus is $x^2 + y^2 = c^2$.

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 the locus of the foot of the perpendicular drawn from the origin to the line is:

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