Question

1. A piece of a broken plate was dug up in an archaeological site. It was put on top of a grid 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 2 describing the boundary of the plate.
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Answer 1

Answer:

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

(x-a)^2+(y-b)^2=R^2(x−a)

2

+(y−b)

2

=R

2

,

where R>0.R>0.

Since the arc of the plate (the circle) passing through A(7,0), B(1,4)A(7,0),B(1,4) and C(7,2)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}

⎩

⎪

⎪

⎨

⎪

⎪

⎧

(7−a)

2

+b

2

=R

2

(1−a)

2

+(4−b)

2

=R

2

(7−a)

2

+(2−b)

2

=R

2

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}

⎩

⎪

⎪

⎨

⎪

⎪

⎧

(7−a)

2

+b

2

=R

2

(1−a)

2

+(4−b)

2

=R

2

(2−b)

2

−b

2

=0

\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}

⎩

⎪

⎪

⎨

⎪

⎪

⎧

(7−a)

2

+b

2

=R

2

(1−a)

2

+(4−b)

2

=R

2

4−4b+b

2

−b

2

=0

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

(7−a)

2

+b

2

=R

2

(1−a)

2

+(4−b)

2

=R

2

b=1

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

(7−a)

2

+1=R

2

(1−a)

2

+9=R

2

b=1

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

(7−a)

2

−(1−a)

2

−8=0

(1−a)

2

+9=R

2

b=1

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

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⎪

⎪

⎨

⎪

⎪

⎧

a

2

−14a+49−1+2a−a

2

−8=0

(1−a)

2

+9=R

2

b=1

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

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⎪

⎪

⎨

⎪

⎪

⎧

−12a+40=0

(1−a)

2

+9=R

2

b=1

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

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⎪

⎪

⎨

⎪

⎪

⎧

a=

3

10

(1−a)

2

+9=R

2

b=1

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

a=

3

10

(−

3

7

)

2

+9=R

2

b=1

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

a=

3

10

9

130

=R

2

b=1

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

a=

3

10

R=

3

130

b=1

The center of the circle (plate) is O(\frac{10}{3},1)O(

3

10

,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}.(x−

3

10

)

2

+(y−1)

2

=

9

130

.

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

(x-a)^2+(y-b)^2=R^2(x−a)

2

+(y−b)

2

=R

2

,

where R>0.R>0.

Since the arc of the plate (the circle) passing through A(7,0), B(1,4)A(7,0),B(1,4) and C(7,2)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}

⎩

⎪

⎪

⎨

⎪

⎪

⎧

(7−a)

2

+b

2

=R

2

(1−a)

2

+(4−b)

2

=R

2

(7−a)

2

+(2−b)

2

=R

2

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}

⎩

⎪

⎪

⎨

⎪

⎪

⎧

(7−a)

2

+b

2

=R

2

(1−a)

2

+(4−b)

2

=R

2

(2−b)

2

−b

2

=0

\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}

⎩

⎪

⎪

⎨

⎪

⎪

⎧

(7−a)

2

+b

2

=R

2

(1−a)

2

+(4−b)

2

=R

2

4−4b+b

2

−b

2

=0

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

(7−a)

2

+b

2

=R

2

(1−a)

2

+(4−b)

2

=R

2

b=1

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

(7−a)

2

+1=R

2

(1−a)

2

+9=R

2

b=1

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

(7−a)

2

−(1−a)

2

−8=0

(1−a)

2

+9=R

2

b=1

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

a

2

−14a+49−1+2a−a

2

−8=0

(1−a)

2

+9=R

2

b=1

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

−12a+40=0

(1−a)

2

+9=R

2

b=1

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

a=

3

10

(1−a)

2

+9=R

2

b=1

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

a=

3

10

(−

3

7

)

2

+9=R

2

b=1

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

a=

3

10

9

130

=R

2

b=1

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

⎩

⎪

⎪

⎨

⎪

⎪

⎧

a=

3

10

R=

3

130

b=1

The center of the circle (plate) is O(\frac{10}{3},1)O(

3

10

,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}.(x−

3

10

)

2

+(y−1)

2

=

9

130

.

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Answer 2

Given:

Arc of the circular plate goes through A(7, 0), B(1, 4), and C(7, 2).

To find:

Center of the circle and general equation of the circle.

Solution:

Standard equation of a circle whose radius R>0 and center O(a, b) is

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

Since the arc of the plate goes through A(7, 0), B(1, 4), and C(7, 2), we can write,

$(7-a)^2+b^2=R^2$-(i)

$(1-a)^2+(4-b)^2=R^2$-(ii)

$(7-a)^2+(2-b)^2=R^2$-(iii)

From (i) we have,

$(7-a)^2=R^2-b^2$

Putting this value in (iii),

$(2-b)^2-b^2=0$

$4+b^2-4b-b^2=0$

⇒$b=1$

Now, putting the value of b in (ii), we get,

$R^2=(1-a)^2+3^2$

⇒$R^2=(1-a)^2+9$

We can put this value of $R^2$ and b in (ii),

$(7-a)^2+1=(1-a)^2+9$

⇒$a^2-14a+49-1-a^2+2a-8=0$

⇒$-12a+40=0$

⇒$a=\frac{10}{3}$

Putting the value of $a$ in (ii), we get,

$R^2=(1-\frac{10}{3} )^2+3^2$

⇒$R^2=(-\frac{7}{3})^2+9$

$R^2=\frac{130}{9}$

Hence, the center of the circle is O$(\frac{10}{3}, 1)$ and the general equation representing the boundary of the plate is $(x-\frac{10}{3})^2+(y-1)^2=\frac{130}{9}$.