Question

4.) the endpoints of a diameter of a circle are located at (-14, -2) and (-8,6).
a. find the coordinates of the center of the circle.
b. find the radius of the circle.
c. write the equation of the circle in standard form.

Explanation:

Part a:

Step1: Recall midpoint formula

The midpoint \( M(x_m, y_m) \) of two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \( x_m=\frac{x_1 + x_2}{2} \), \( y_m=\frac{y_1 + y_2}{2} \). Here, \( (x_1,y_1)=(- 14,-2) \) and \( (x_2,y_2)=(-8,6) \).

Step2: Calculate x - coordinate of center

\( x_m=\frac{-14+( - 8)}{2}=\frac{-22}{2}=-11 \)

Step3: Calculate y - coordinate of center

\( y_m=\frac{-2 + 6}{2}=\frac{4}{2}=2 \)

Step1: Recall distance formula

The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \). The radius \( r \) is half of the diameter. First, find the length of the diameter (distance between \((-14,-2)\) and \((-8,6)\)) and then divide by 2.

Step2: Calculate the diameter length

\( d=\sqrt{(-8-( - 14))^2+(6-( - 2))^2}=\sqrt{(6)^2+(8)^2}=\sqrt{36 + 64}=\sqrt{100} = 10 \)

Step3: Calculate the radius

Since \( r=\frac{d}{2} \), \( r=\frac{10}{2}=5 \)

Step1: Recall standard form of circle equation

The standard form of the equation of a circle with center \((h,k)\) and radius \( r \) is \( (x - h)^2+(y - k)^2=r^2 \). We know \( h=-11 \), \( k = 2 \) and \( r = 5 \).

Step2: Substitute values into the formula

Substitute \( h=-11 \), \( k = 2 \) and \( r = 5 \) into \( (x - h)^2+(y - k)^2=r^2 \). We get \( (x-(-11))^2+(y - 2)^2=5^2 \), which simplifies to \( (x + 11)^2+(y - 2)^2=25 \)

Answer:

The coordinates of the center are \((-11,2)\)

Part b: