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(A) Let the line passing through $(a, b)$ be $y - b = m(x - a)$,where $m$ is the slope.This can be written as $mx - y + (b - ma) = 0$.Let $(h, k)$ be

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$x^2 + y^2 - ax - by = 0$

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(A) Let the line passing through $(a, b)$ be $y - b = m(x - a)$,where $m$ is the slope.This can be written as $mx - y + (b - ma) = 0$.Let $(h, k)$ be the foot of the perpendicular from the origin $(0, 0)$ to this line.The slope of the line joining $(0, 0)$ and $(h, k)$ is $k/h$.Since the line is perpendicular to the given line,the product of their slopes is $-1$: $(m)(k/h) = -1 \implies m = -h/k$.Substituting $m = -h/k$ into the line equation: $(-h/k)x - y + (b - (-h/k)a) = 0$.Multiplying by $k$: $-hx - ky + bk + ah = 0 \implies hx + ky = ah + bk$.Since $(h, k)$ lies on the line $mx - y + b - ma = 0$,we have $mh - k + b - ma = 0$.Substituting $m = -h/k$: $(-h/k)h - k + b - (-h/k)a = 0$.$-h^2 - k^2 + bk + ah = 0 \implies h^2 + k^2 - ah - bk = 0$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - ax - by = 0$.

The equation of the locus of the foot of the perpendicular drawn from the origin to a line passing through a fixed point $(a, b)$ is:

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