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(A) Let $P(\alpha, \beta)$ be the given fixed point and $O(0, 0)$ be the origin.Let $Q(x, y)$ be the foot of the perpendicular from the origin $O$ to

Vedclass · Accepted Answer

$x^{2}+y^{2}-\alpha x-\beta y=0$

Vedclass · Answer

(A) Let $P(\alpha, \beta)$ be the given fixed point and $O(0, 0)$ be the origin.Let $Q(x, y)$ be the foot of the perpendicular from the origin $O$ to the variable line passing through $P$.Since $OQ \perp PQ$,the angle $\angle OQP = 90^{\circ}$.This implies that the point $Q$ lies on a circle with $OP$ as its diameter.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.Substituting the points $O(0, 0)$ and $P(\alpha, \beta)$:$(x-0)(x-\alpha) + (y-0)(y-\beta) = 0$$x(x-\alpha) + y(y-\beta) = 0$$x^{2} - \alpha x + y^{2} - \beta y = 0$Thus,the locus is $x^{2} + y^{2} - \alpha x - \beta y = 0$.

$A$ 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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