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A ball rests in the corner of a room where two walls and the floor meet…

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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 ball rests in the corner of a room where two walls and the floor meet. Take the corner as the origin, with the floor as the $xy$ -plane and the walls as the other two coordinate planes. All lengths are in metres. The ball touches all three of these surfaces, so its centre is at $(r, r, r)$ for its radius $r$ . A small hook is fixed at the point $(1, 2, 2)$ , and the ball also touches the hook, so the hook is on its surface. Write the equation that $r$ must satisfy. There are two balls that fit. Find the larger radius and the smaller radius, each to two decimal places. Do not round before the last step.

Need a reminder? The key ideaShow

A sphere with centre $(a, b, c)$ and radius $r$ is $(x - a)^2 + (y - b)^2 + (z - c)^2 = r^2$ . In the general form $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$ the centre is $(-u, -v, -w)$ and $r^2 = u^2 + v^2 + w^2 - d$ . The plane $z = h$ cuts a circle of radius $\sqrt{r^2 - (h - c)^2}$ .

The worked solution

Have you tried it?

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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