Question

Write the equation of a circle that has diameter endpoints (16, -4) and (-12,4)

Answer 1

The equation of the circle with diameter endpoints (16, -4) and (-12, 4) is
`(x - 1)^2 + y^2 = 196.`

Step-by-step explanation:

To write the equation of a circle, we need to know the coordinates of the center and the radius. In this case, we have the diameter endpoints, which can be used to find the center. The midpoint of the diameter is the center, and the distance between the center and one of the endpoints is the radius.

Using the coordinates (16, -4) and (-12, 4), we can find the midpoint by averaging the x-coordinates and the y-coordinates: (16 + (-12))/2 = 2/2 = 1 and (-4 + 4)/2 = 0/2 = 0.

So, the coordinates of the center are (1, 0). The radius is half the distance between the two endpoints of the diameter, which is ((-12) - 16)/2 = (-28)/2 = -14.

Therefore, the equation of the circle isx
`(x - 1)^2 + y^2 = 196.`

Answer 2

Answer::

`(x-2)^2+y^2=212`

Explanation:

Given the endpoints of the diameter of a circle, we are required to write the equation of the circle.

The standard form of the equation of a circle is given as:

`(x-a)^2+(y-b)^2=r^2\text{ where }\begin{cases}Center=(a,b) \\ Radius=r\end{cases}`

So, we need to find two things, the center, and the radius.

Center

The center of the circle is the midpoint of the diameter.

Given the endpoints as: (16, -4) and (-12,4)

`\text{Midpoint}=((16+(-12))/(2),(-4+4)/(2))=((4)/(2),(0)/(2))=(2,0)`

The center, (a,b)=(2,0).

Radius

The radius is the distance between the center, (2,0) and one of the endpoints of the diameter.

Using the distance formula, we find the distance between (2,0) and (16,-4).

`\begin{gathered} Distance=√((x_2-x_1)^2+(y_2-y_1)^2) \\ =\sqrt[]{(16-2)^2+(-4-0)^2}=\sqrt[]{(14)^2+(-4)^2}=\sqrt[]{212} \\ \implies\text{Radius}=√(212) \end{gathered}`

Substitute these values into the equation given earlier:

`\begin{gathered} (x-2)^2+(y-0)^2=\sqrt[]{212}^2\text{ where }\begin{cases}Center,(a,b)=(2,0) \\ Radius,r=\sqrt[]{212}\end{cases} \\ (x-2)^2+y^2=212 \end{gathered}`

The equation of the circle is:

`(x-2)^2+y^2=212`