Question

identify the center and write the equation of the circle that has the ends of the diameter at (17, 12) and (13, 16)

Answer 1

The Solution:

Given the ends of a diameter:

`(17,12)\text{ and }(13,16)`

Required:

To write the equation of the circle.

Step 1:

Find the center of the circle by using the midpoint formula.

`Midpoint=((x_1+x_2)/(2),(y_1+y_2)/(2))`

In this case,

`\begin{gathered} x_1=17 \\ y_1=12 \\ x_2=13 \\ y_2=16 \end{gathered}`

Substituting, we get

`\begin{gathered} \text{ Midpoint}=((17+13)/(2),(12+16)/(2))=((30)/(2),(28)/(2))=(15,14) \\ So, \\ The\text{ center}=(15,14) \end{gathered}`

Step 2:

Find the radius of the circle.

Using the distance between two points formula:

`r^2=(x_2-x_1)^2+(y_2-y_1)^2`
`\begin{gathered} r^2=(13-17)^2+(16-12)^2 \\ \\ r^2=(-4)^2+(4)^2 \\ \\ r^2=16+16 \\ \\ r^2=32 \end{gathered}`

Step 3:

Write the equation of the circle.

`\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ \text{ Where} \\ h=15 \\ k=14 \\ r=32 \end{gathered}`
`\begin{gathered} (x-15)^2+(y-14)^2=32^2 \\ \\ (x-15)^2+(y-14)^2-1024=0 \end{gathered}`

Therefore, the correct answers are:

`\begin{gathered} center=(15,14) \\ \\ Equation:\text{ }(x-15)^2+(y-14)^2-1024=0 \end{gathered}`