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A variable plane passes through a fixed point $(a,b,c)$ and cuts the coordinate axes at $P,Q,R$. Then the coordinates $(x,y,z)$ of the centre of the sphere passing through $P,Q,R$ and the origin satisfy which of the following equation?
(A) $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$
(B) $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=3$
(C) $ax+by+cz=1$
(D) $ax+by+cz=a^2+b^2+c^2$
I haven't been able to make much progress in this question. We can take the equation of the plane to be $a_1(x-a)+b_1(y-b)+c_1(z-c)=0$. Then we can find $P,Q,R$. Then I am not sure what to do next.
P.S.- I don't know the general equation of a sphere.

idpd15
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  • Note that the sphere cuts the three planes $x=0;y=0; z=0$ in circles in which triangles $OPR; OQR; OPQ$ are inscribed. These are right angled triangles, so the centres of the circles are at the midpoints of the respective hypotenuses. Take a specific circle - the centre of the sphere is "above" it in the sense that it lies on the perpendicular to the plane which meets it at the centre of the circle. This makes it easy to locate the centre of the sphere - no quadratics required - just systematic working through. It might help to draw a diagram. – Mark Bennet Apr 03 '14 at 11:20
  • @MarkBennet Ok, yeah thanks it works :) – idpd15 Apr 03 '14 at 11:43

1 Answers1

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To follow my comment:

Your plane is $Ax+By+Cz=Aa+Bb+Cc=D$ - to check, see that the given point lies on the plane, and that this is a linear form.

It cuts the co-ordinate axes at $(\frac DA,0,0); (0,\frac DB,0); (0,0,\frac DC)$

The centre of the circle at which the sphere cuts the plane $z=0$ is half way along the line between $(\frac DA,0,0)$ and $(0,\frac DB,0)$ i.e. at $(\frac {D}{2A},\frac {D}{2B},0)$.

The perpendicular to the plane at this point is (easy because it is a co-ordinate plane) $x=\frac {D}{2A}; y=\frac {D}{2B}$

You should be able to use this method to work out the co-ordinates of the centre of the sphere. Then check which equation the centre will satisfy.

Mark Bennet
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    There is just 1 problem with your solution. That is, you have taken a specific plane and NOT any plane through the point. We can just replace your equation of the plane with a general equation and it will work. Thanks anyways! – idpd15 Apr 03 '14 at 11:49
  • @MridulSachdeva Have adapted answer to be a full answer, having concentrated on the second half, and not done the general case. – Mark Bennet Apr 03 '14 at 14:27