Question

Ends of a diameter: (16,-2) and (16,-4). Find the standard equation of the circle.

Answer 1

Given the endpoints of the diameter are (16,-2) and (16,-4).

Find the center of the circle by finding the midpoint of the line segment joining (16, -2) and (16,-4).

`\begin{gathered} M=((x_1+x_2)/(2),(y_1+y_2)/(2)) \\ =((16+16)/(2),(-2-4)/(2)) \\ =(16,-3) \end{gathered}`

Now, find the radius of the circle using the distance formula.

`\begin{gathered} d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ =\sqrt[]{(16-16)^2+(-4-(-2))^2} \\ =\sqrt[]{(-2)^2} \\ =2 \end{gathered}`

Thus, the diameter of the circle is 2. Hence the radius is 1 unit.

The standard equation of a circle is

`(x-h)^2+(y-k)^2=r^2`

where (h, k) is the center and r is the radius.

Substitute the obtained values into the equation.

`(x-16)^2+(y+3)^2=1`