3520.11 – The Balancing Pot

Here is a pot, balanced on a pair of steel slabs that meet at right angles below the pot. The intersection of the slabs is treated here as the origin of a set of co-ordinate axes which hit the edge of the pot at the distances −6,15,-6, 15, and 88 as shown.

What is the diameter of the pot?

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Solution

The problem gives, in effect. three points on a circle as in the figure below. As three points determine a circle, it should be possible to find the equation of this one:

(x−h)2+(y−k)2=r2,(x-h)^2 + (y-k)^2 = r^2,

where (h,k)(h, k) is the location of the center, and so determine the radius rr. Plugging in the three points gives the three equations:

36+12h+h2+k2=r2,225−30h+h2+k2=r2,h2+64−16k+k2=r2. \begin{aligned} 36 + 12 h + h^2 + k^2 &=& r^2, \\ 225 - 30 h + h^2 + k^2 &=& r^2, \\ h^2 + 64 - 16 k + k^2 &=& r^2. \end{aligned}

Subtracting the equations in pairs gives

42h−189=0,36+12h−64−16k=0. \begin{aligned} 42 h - 189 &=& 0, \\ 36 + 12 h - 64 - 16k &=& 0. \end{aligned}

These two linear equations are easily solved. It turns out that h=189/42=9/2h = 189/42 = 9/2 and k=−26/16=−13/8k = -26/16 = -13/8. Plugging this information into any of the first three equations, we discover that r=85/4=21.25.r = 85/4 = 21.25.

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