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A plane passes through a fixed point $(a,b,c)$ and cuts the axes at the points $A,B,C$ respectively. Find the locus of the centre of the sphere which passes through the origin $O$ and $A,B,C$.
asked Nov 11, 2017 by randomisation (1,920 points)  

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Any sphere passing through origin is $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz = 0$ where $(-u,-v,-w)$ is the centre of the sphere.

At the point where it cuts the x-axis, $y=0, z=0$

$\implies x^2 + 2ux = 0 \implies x = -2u$

Thus, point of intersection is $(-2u,0,0)$. Similarly, points of intersection with other two axes are $(0,-2v,0)$ and $(0,0,-2w)$

Plane passing through these given points $A,B,C$ is $\frac{x}{-2u} + \frac{y}{-2v} + \frac{z}{-2w} = 1$

It passes through $(a,b,c)$, thus, $\frac{a}{-2u} + \frac{b}{-2v} + \frac{c}{-2w} = 1$

As the centre of sphere is $(-u,-v,-w)$, the locus is:

$\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 2$
answered Nov 11, 2017 by randomisation (1,920 points)  
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