Question

Write the equation of the circle given the following information.Endpoints of the diameter are at (-9,10) and (-1,12)

Answer 1

Given the points:

(-9, 10) and (-1, 12)

Let's find the equation of the circle using the points which are the endpoints of the diameter.

Apply the general equation for a circle:

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

Where:

• (h, k) is the center

,

• r is the radius of the circle.

Let's first find the diameter of the circle using the distance between points formula:

`d=√((x2-x1)^2+(y2-y1)^2)`

Where:

(x1, y1) ==> (-9, 10)

(x2, y2) ==> (-1, 12)

Hence, we have:

`\begin{gathered} d=√((-1-(-9))^2+(12-10)^2) \\ \\ d=√((-1+9)^2+(12-10)^2) \\ \\ d=√((8)^2+(2)^2) \\ \\ d=√(64+4) \\ \\ d=√(68) \\ \\ d=8.2 \end{gathered}`

The diameter of the circle is 8.2 units.

To find the radius, we have:

radius = diameter/2 = 8.2/2 = 4.1 units

The radius of the circle is 4.1 units.

Now, let's find the center of the circle.

To find the center of the circle, apply the midpoint formula:

`m=((x1+x2))/(2),((y1+y2))/(2)`

Thus, we have:

`\begin{gathered} (h,k)=(-9+(-1))/(2),(10+12)/(2) \\ \\ (h,k)=(-9-1)/(2),(10+12)/(2) \\ \\ (h,k)=(-10)/(2),(22)/(2) \\ \\ (h,k)=(-5,11) \end{gathered}`

The center of the circle is (-5, 11).

Therefore, the equation of the circle with the points is:

`\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ \\ (x-(-5))^2+(y-11)^2=4.1^2 \\ \\ (x+5)^2+(y-11)^2=17 \end{gathered}`

ANSWER:

`(x+5)^2+(y-11)^2=17`