If frac{x}{alpha} + frac{y}{beta} = 1 touches | vedclass

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(B) The condition for the line $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ to touch the circle $x^2 + y^2 = a^2$ is that the perpendicular distance from

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(B) The condition for the line $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ to touch the circle $x^2 + y^2 = a^2$ is that the perpendicular distance from the center $(0, 0)$ to the line must be equal to the radius $a$.The equation of the line can be written as $\frac{1}{\alpha}x + \frac{1}{\beta}y - 1 = 0$.The perpendicular distance $d$ is given by $d = \frac{|\frac{1}{\alpha}(0) + \frac{1}{\beta}(0) - 1|}{\sqrt{(\frac{1}{\alpha})^2 + (1/\beta)^2}} = a$.Squaring both sides,we get $\frac{1}{(\frac{1}{\alpha})^2 + (1/\beta)^2} = a^2$.This simplifies to $(\frac{1}{\alpha})^2 + (1/\beta)^2 = \frac{1}{a^2}$.Let $X = \frac{1}{\alpha}$ and $Y = \frac{1}{\beta}$. The equation becomes $X^2 + Y^2 = (\frac{1}{a})^2$,which represents a circle.

If $\frac{x}{\alpha} + \frac{y}{\beta} = 1$ touches the circle $x^2 + y^2 = a^2$,then the point $(\frac{1}{\alpha}, \frac{1}{\beta})$ lies on a/an

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