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(B) Let the centre of the circle be $C(h, k)$. Since the circle passes through the origin $(0,0)$,its radius $r$ is given by $r^2 = h^2 + k^2$.The equ

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$y^2+6x=13$

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(B) Let the centre of the circle be $C(h, k)$. Since the circle passes through the origin $(0,0)$,its radius $r$ is given by $r^2 = h^2 + k^2$.The equation of the circle is $(x-h)^2 + (y-k)^2 = h^2 + k^2$,which simplifies to $x^2 + y^2 - 2hx - 2ky = 0$.The perpendicular distance $d$ from the centre $(h, k)$ to the line $x=3$ (or $x-3=0$) is $d = |h-3|$.In the right-angled triangle formed by the centre,the midpoint of the chord,and an endpoint of the chord,we have $r^2 = d^2 + (L/2)^2$,where $L=4$ is the length of the chord.Thus,$h^2 + k^2 = (h-3)^2 + 2^2$.$h^2 + k^2 = h^2 - 6h + 9 + 4$.$k^2 = -6h + 13$.Replacing $(h, k)$ with $(x, y)$,the locus is $y^2 = -6x + 13$,or $y^2 + 6x = 13$.

The locus of the centre of a circle which passes through the origin and cuts off a chord of length $4$ units from the line $x=3$ is

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