Circle Equations
Generally, the equation of a circle is organized like this: [tex](x-h)^2+(y-k)^2=r^2[/tex], where (h,k) is the centre and r is the radius.
Solving the Question
We're given the endpoints of the diameter: [tex](-6,-3)[/tex] and [tex](-6,-8)[/tex].
First, we can find the centre of the circle by finding the midpoint of these two endpoints.
[tex]midpoint=(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})[/tex]
Plug in the points:
[tex]midpoint=(\dfrac{-6+-6}{2},\dfrac{-3+-8}{2})\\\\midpoint=(-6,\dfrac{-11}{2})[/tex]
Therefore, the centre of the circle is [tex](-6,\dfrac{-11}{2})[/tex]. Plug this into the general equation:
[tex](x-h)^2+(y-k)^2=r^2\\\\(x-(-6))^2+(y-(-\dfrac{11}{2}))^2=r^2\\\\(x+6)^2+(y+\dfrac{11}{2})^2=r^2[/tex]
To find the radius of the circle, we can calculate the distance between the centre and one of the given endpoints:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2[/tex]
Plug in the centre and one of the endpoints:
[tex]d=\sqrt{(-6-x_1)^2+(\dfrac{-11}{2}-y_1)^2}\\d=\sqrt{(-6-(-6))^2+(\dfrac{-11}{2}-(-3))^2}\\d=\sqrt{(-6+6)^2+(\dfrac{-11}{2}+3)^2}\\\\d=\sqrt{(0)^2+(-5.5+3)^2}\\\\d=\sqrt{(0)^2+(-2.5)^2}\\\\d=\sqrt{(0)^2+(-\dfrac{5}{2})^2}\\\\d=\sqrt{0+\dfrac{25}{4}}\\d=\sqrt{\dfrac{25}{4}}\\\\d=\dfrac{5}{2}[/tex]
Therefore, the radius of the circle is [tex]\dfrac{5}{2}[/tex]. Plug this into the general equation:
[tex](x+6)^2+(y+\dfrac{11}{2})^2=r^2\\\\(x+6)^2+(y+\dfrac{11}{2})^2=(\dfrac{5}{2})^2\\\\(x+6)^2+(y+\dfrac{11}{2})^2=\dfrac{25}{4}[/tex]
Answer
[tex](x+6)^2+(y+\dfrac{11}{2})^2=\dfrac{25}{4}[/tex]
