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A tower A stands at the origin and a tower B stands at (6, 0, 0), with distances in metres…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A tower $A$ stands at the origin and a tower $B$ stands at $(6, 0, 0)$ , with distances in metres. A buoy $P$ floats so that its distance from $A$ is always exactly twice its distance from $B$ . The buoy can be at many places, and together they form a sphere. Write the condition $PA^2 = 4PB^2$ and simplify it to the standard form of a sphere. Find the $x$ -coordinate of the centre and the radius of the sphere. Then find where the sphere cuts the $x$ -axis: give the smaller $x$ and the larger $x$ .

Need a reminder? The key ideaShow

For $PA = PB$ , square both sides. The squares of $x$ , $y$ and $z$ cancel and a plane is left. A fixed ratio $PA = \lambda PB$ ( $\lambda \ne 1$ ) gives a sphere. A point equally far from four points that are not in one plane comes from three linear equations.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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