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(C) Let the foot of the perpendicular from the origin $(0,0)$ to the variable line $\frac{x}{a} + \frac{y}{b} = 1$ be $P(x_1, y_1)$.Since $P(x_1, y_1)

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a circle of radius $= 2$

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(C) Let the foot of the perpendicular from the origin $(0,0)$ to the variable line $\frac{x}{a} + \frac{y}{b} = 1$ be $P(x_1, y_1)$.Since $P(x_1, y_1)$ lies on the line $\frac{x}{a} + \frac{y}{b} = 1$,we have $\frac{x_1}{a} + \frac{y_1}{b} = 1$.The slope of the line $\frac{x}{a} + \frac{y}{b} = 1$ is $m_1 = -\frac{b}{a}$.The slope of the line segment $OP$ is $m_2 = \frac{y_1}{x_1}$.Since $OP$ is perpendicular to the line,$m_1 	imes m_2 = -1$,so $(-\frac{b}{a}) 	imes (\frac{y_1}{x_1}) = -1$,which implies $\frac{y_1}{x_1} = \frac{a}{b}$,or $\frac{x_1}{a} = \frac{y_1}{b} = k$.Substituting this into the line equation: $k + k = 1 \Rightarrow k = \frac{1}{2}$.Thus,$x_1 = \frac{a}{2}$ and $y_1 = \frac{b}{2}$,which means $a = 2x_1$ and $b = 2y_1$.Given $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$,we substitute $a$ and $b$: $\frac{1}{(2x_1)^2} + \frac{1}{(2y_1)^2} = \frac{1}{4}$.$\frac{1}{4x_1^2} + \frac{1}{4y_1^2} = \frac{1}{4}$ $\Rightarrow \frac{1}{x_1^2} + \frac{1}{y_1^2} = 1$ $\Rightarrow x_1^2 + y_1^2 = x_1^2 y_1^2$.Alternatively,using the perpendicular distance formula $d = \frac{|c|}{\sqrt{A^2 + B^2}}$ from $(0,0)$ to $\frac{x}{a} + \frac{y}{b} - 1 = 0$:$d = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} = \sqrt{x_1^2 + y_1^2}$.Since $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$,we have $\sqrt{x_1^2 + y_1^2} = \sqrt{4} = 2$.Thus,$x_1^2 + y_1^2 = 4$,which is a circle with radius $2$.

Let $a$ and $b$ be any two numbers satisfying $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$. Then,the foot of the perpendicular from the origin on the variable line $\frac{x}{a} + \frac{y}{b} = 1$ lies on

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