Question

Find the standard form of the equation for the circle with the following properties. Endpoints of a diameter are (-6,5) and (-2,-7)

Answer 1

For the question given, we are told to find the equation of a circle. To derive the equation we will follow the steps below.

Step 1:

Get the center of the circle

The center of the circle is the midpoint of the endpoints given

so that

`\begin{gathered} c=(-6-2)/(2),(5-7)/(2)=(-8)/(2),(-2)/(2)=-4,-1 \\ \\ c=-4,-1 \end{gathered}`

Thus the center of the circle is (-4,-1)

Step 2: Find the radius of the circle

To do this, we will get the distance between the two points and then divide it by 2

`\begin{gathered} \text{The distance betwe}en\text{ two points is given by} \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \end{gathered}`

`\begin{gathered} d=\sqrt[]{(-2+6)^2+(-7-5)^2} \\ d=\sqrt[]{4^2+(-12)^2} \\ d=\sqrt[]{16+144} \\ d=\sqrt[]{160} \\ d=4\sqrt[]{10} \end{gathered}`

Then, the radius of the circle is

`r=(d)/(2)=\frac{4\sqrt[]{10}}{2}=2\sqrt[]{10}`

Step 3: List the parameters and apply the equation of the circle

`(x-a)^2+(y-b)^2=r^2`

Since the center of the circle is (-4,-1) and the radius of the circle is 2√10

Then

`\begin{gathered} a=-4 \\ b=-1 \end{gathered}`

`r=2\sqrt[]{10}`

Step 4. Find the equation of the circle

`\begin{gathered} (x-(-4))^2+(y-(-1))^2=(2\sqrt[]{10})^2 \\ (x+4)^2+(y+1)^2=(4*10) \\ x^2+8x+16+y^2+2y+1=40 \\ x^2+8x+y^2+2y+17=40 \\ x^2+8x+y^2+2y=40-17 \\ x^2+8x+y^2+2y=23 \end{gathered}`

Then the equation of the circle is:

`x^2+8x+y^2+2y=23`