A line through (0,0) cuts the circle x^2 + y^ | vedclass

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(D) Let the line passing through the origin $(0,0)$ be $y = mx$,or $mx - y = 0$. Let $(h, k)$ be the centre of the circle drawn with $AB$ as the diame

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

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(D) Let the line passing through the origin $(0,0)$ be $y = mx$,or $mx - y = 0$. Let $(h, k)$ be the centre of the circle drawn with $AB$ as the diameter. The centre of the circle $x^2 + y^2 - 2ax = 0$ is $C(a, 0)$. Since the line $AB$ is a chord of the circle,the perpendicular from the centre $C(a, 0)$ to the chord $AB$ bisects the chord at its midpoint $M(h, k)$. Thus,the line $CM$ is perpendicular to the line $AB$. The slope of $AB$ is $m = \frac{k}{h}$. The slope of $CM$ is $\frac{k - 0}{h - a} = \frac{k}{h - a}$. Since $CM \perp AB$,the product of their slopes is $-1$: $\left(\frac{k}{h}ight) 	imes \left(\frac{k}{h - a}ight) = -1$ $k^2 = -h(h - a)$ $k^2 = -h^2 + ah$ $h^2 + k^2 - ah = 0$ Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 + y^2 - ax = 0$.

$A$ line through $(0,0)$ cuts the circle $x^2 + y^2 - 2ax = 0$ at points $A$ and $B$. The locus of the centre of the circle drawn on $AB$ as a diameter is:

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