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A variable line passes through the fixed point $(\alpha, \beta) .$ The locus of the foot of the perpendicular from the origin on the line is
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Verified Answer
The correct answer is:
$x^{2}+y^{2}-\alpha x-\beta y=0$
Let $(\alpha, \beta)$ be the given point, let $Q(x, y)$ be the foot of the perpendicular, and let $O$ be the origin. The line can have any direction
$$
\angle P Q O=90^{\circ}
$$
Point $Q$ lies on the circle having diameter $O P$.
The locus of point $Q$.
$$
(x-0)(x-\alpha)+(y-0)(y-\beta)=0
$$
$\Rightarrow \quad x^{2}-x \alpha+y^{2}-y \beta=0$
$$
\Rightarrow \quad x^{2}+y^{2}-\alpha x-\beta y=0
$$
$$
\angle P Q O=90^{\circ}
$$
Point $Q$ lies on the circle having diameter $O P$.
The locus of point $Q$.
$$
(x-0)(x-\alpha)+(y-0)(y-\beta)=0
$$
$\Rightarrow \quad x^{2}-x \alpha+y^{2}-y \beta=0$
$$
\Rightarrow \quad x^{2}+y^{2}-\alpha x-\beta y=0
$$
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