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

1 Answer
May 1, 2016

Answer:

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]}