Question

How do you write the equation for a circle with Center of circle (8,4) radius with endpoint (0,4)?

Answer 1

The answer is: `(x-8)^2+(y-4)^2=64`

Explanation:

You can find the reason for this based on the standard equation of a circle, which is: `(x-h)^2+(y-k)^2=r^2`, where `(h, k)` represents the coordinates of the center of the circle and `r^2` the radius of the circle squared.

Using the standard form, we can deconstruct the given information as such:

• `(h, k)` of the circle is `(8, 4)`, since `(8, 4)` represents the center of the circle.
• The radius of the circle is 8. We know this because a circle is defined as the set of all point equidistant from the center of the circle, and, if one of the endpoints is at `(0, 4)`, then the distance between it and the center `(8, 4)` is 8 (there's no change in y-values). Thus, we can plug each of the values into the standard equation:

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

`(x-(8))^2+(y-(4))^2= (8)^2`

`(x-8)^2+(y-4)^2=64`

*Note that you would have had to use the distance formula if the y-coordinate was not the same.

Graphically:

graph{(x-8)^2+(y-4)^2=64 [-8.78, 27.27, -5.02, 13]}