Question

Give the equation of the circle that has a diameter with endpoints G(-6, — 3) and
R(-6,-8).

Answer 1

Circle Equations

Generally, the equation of a circle is organized like this:
`(x-h)^2+(y-k)^2=r^2`, where (h,k) is the centre and r is the radius.

Solving the Question

We're given the endpoints of the diameter:
`(-6,-3)` and
`(-6,-8)`.

First, we can find the centre of the circle by finding the midpoint of these two endpoints.

`midpoint=((x_1+x_2)/(2),(y_1+y_2)/(2))`

Plug in the points:

`midpoint=((-6+-6)/(2),(-3+-8)/(2))\\\\midpoint=(-6,(-11)/(2))`

Therefore, the centre of the circle is
`(-6,(-11)/(2))`. Plug this into the general equation:

`(x-h)^2+(y-k)^2=r^2\\\\(x-(-6))^2+(y-(-(11)/(2)))^2=r^2\\\\(x+6)^2+(y+(11)/(2))^2=r^2`

To find the radius of the circle, we can calculate the distance between the centre and one of the given endpoints:

`d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2`

Plug in the centre and one of the endpoints:

`d=\sqrt{(-6-x_1)^2+((-11)/(2)-y_1)^2}\\d=\sqrt{(-6-(-6))^2+((-11)/(2)-(-3))^2}\\d=\sqrt{(-6+6)^2+((-11)/(2)+3)^2}\\\\d=√((0)^2+(-5.5+3)^2)\\\\d=√((0)^2+(-2.5)^2)\\\\d=\sqrt{(0)^2+(-(5)/(2))^2}\\\\d=\sqrt{0+(25)/(4)}\\d=\sqrt{(25)/(4)}\\\\d=(5)/(2)`

Therefore, the radius of the circle is
`(5)/(2)`. Plug this into the general equation:

`(x+6)^2+(y+(11)/(2))^2=r^2\\\\(x+6)^2+(y+(11)/(2))^2=((5)/(2))^2\\\\(x+6)^2+(y+(11)/(2))^2=(25)/(4)`

Answer

`(x+6)^2+(y+(11)/(2))^2=(25)/(4)`