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

$\begin{gathered} c=\frac{-6-2}{2},\frac{5-7}{2}=\frac{-8}{2},\frac{-2}{2}=-4,-1 \\ \\ c=-4,-1 \end{gathered}$

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

$r=\frac{d}{2}=\frac{4\sqrt[]{10}}{2}=2\sqrt[]{10}$

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

$\begin{gathered} (x-(-4))^2+(y-(-1))^2=(2\sqrt[]{10})^2 \\ (x+4)^2+(y+1)^2=(4\times10) \\ x^2+8x+16+y^2+2y+1=40 \\ x^2+8x+y^2+2y+17=40 \\ x^2+8x+y^2+2y=40-17 \\ x^2+8x+y^2+2y=23 \end{gathered}$

Then the equation of the circle is:

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