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

May 1, 2016

The answer is: ${\left(x - 8\right)}^{2} + {\left(y - 4\right)}^{2} = 64$

#### Explanation:

You can find the reason for this based on the standard equation of a circle, which is: ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$, where $\left(h , k\right)$ 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:

• $\left(h , k\right)$ of the circle is $\left(8 , 4\right)$, since $\left(8 , 4\right)$ 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 $\left(0 , 4\right)$, then the distance between it and the center $\left(8 , 4\right)$ is 8 (there's no change in y-values). Thus, we can plug each of the values into the standard equation:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

${\left(x - \left(8\right)\right)}^{2} + {\left(y - \left(4\right)\right)}^{2} = {\left(8\right)}^{2}$

${\left(x - 8\right)}^{2} + {\left(y - 4\right)}^{2} = 64$

*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]}