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

Identify The Center And Write The Equation Of The Circle That Has The Ends Of The Diameter At 17 12 And 13 16 class=

Sagot :

The Solution:

Given the ends of a diameter:

[tex](17,12)\text{ and }(13,16)[/tex]

Required:

To write the equation of the circle.

Step 1:

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

[tex]Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})[/tex]

In this case,

[tex]\begin{gathered} x_1=17 \\ y_1=12 \\ x_2=13 \\ y_2=16 \end{gathered}[/tex]

Substituting, we get

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

Step 2:

Find the radius of the circle.

Using the distance between two points formula:

[tex]r^2=(x_2-x_1)^2+(y_2-y_1)^2[/tex][tex]\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}[/tex]

Step 3:

Write the equation of the circle.

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

Therefore, the correct answers are:

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

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