Answer to Question #143334 in Calculus for John

Question

A piece of a broken plate was dug up in an archaeological site. It was put on top of a grid, as shown in Figure 1.15, with the arc of the plate passing through A(7,0), B(1,4) and C(7,2). Find its center, and the standard equation of the circle describing the boundary of the plate.

Answer 1

The standard equation of the circle with center $O(a,b)$ and radius $R$ is the following:

$(x-a)^2+(y-b)^2=R^2$,

where $R>0.$

Since the arc of the plate (the circle) passing through $A(7,0), B(1,4)$ and $C(7,2)$, we have the following system:

$\begin{cases}(7-a)^2+b^2=R^2\\(1-a)^2+(4-b)^2=R^2\\(7-a)^2+(2-b)^2=R^2\end{cases}$

which is equivalent to

$\begin{cases}(7-a)^2+b^2=R^2\\(1-a)^2+(4-b)^2=R^2\\(2-b)^2-b^2=0\end{cases}$

$\begin{cases}(7-a)^2+b^2=R^2\\(1-a)^2+(4-b)^2=R^2\\4-4b+b^2-b^2=0\end{cases}$

$\begin{cases}(7-a)^2+b^2=R^2\\(1-a)^2+(4-b)^2=R^2\\b=1\end{cases}$

$\begin{cases}(7-a)^2+1=R^2\\(1-a)^2+9=R^2\\b=1\end{cases}$

$\begin{cases}(7-a)^2-(1-a)^2-8=0\\(1-a)^2+9=R^2\\b=1\end{cases}$

$\begin{cases}a^2-14a+49-1+2a-a^2-8=0\\(1-a)^2+9=R^2\\b=1\end{cases}$

$\begin{cases}-12a+40=0\\(1-a)^2+9=R^2\\b=1\end{cases}$

$\begin{cases}a=\frac{10}{3}\\(1-a)^2+9=R^2\\b=1\end{cases}$

$\begin{cases}a=\frac{10}{3}\\(-\frac{7}{3})^2+9=R^2\\b=1\end{cases}$

$\begin{cases}a=\frac{10}{3}\\\frac{130}{9}=R^2\\b=1\end{cases}$

$\begin{cases}a=\frac{10}{3}\\R=\frac{\sqrt{130}}{3}\\b=1\end{cases}$

The center of the circle (plate) is $O(\frac{10}{3},1)$ and the standard equation of the circle describing the boundary of the plate is the following:

$(x-\frac{10}{3})^2+(y-1)^2=\frac{130}{9}.$