Equation of a Circle
Circular equations are organized like this:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
- [tex](h,k)[/tex] is the center of the circle
- [tex]r[/tex] is the radius
Solving the Question
We're given:
- (6, 0) and (-6, 0) are the endpoints of the diameter of a circle
To find the equation of this circle, we must determine two things:
- The center of the circle
- The radius
To determine the radius of the circle, find the distance between the given endpoints and divide by 2:
(6, 0) and (-6, 0)
⇒ There are 6 units between (0,0) and (6,0).
⇒ There are 6 units between (0,0) and (-6,0).
⇒ 6 + 6 = 12; the diameter of the circle is 12 units
⇒ 12 / 2 = 6
Therefore, the radius of the circle is 6 units.
To determine the center of the circle, find the midpoint of the given endpoints:
(6, 0) and (-6, 0)
⇒ The midpoint occurs at the very middle of these two points. The distance between the midpoint and (6,0) is equal to the distance between the midpoint and (-6,0).
⇒ We identified earlier that there are 6 units between (0,0) and (6,0), and that there are 6 units between (0,0) and (-6,0).
Therefore, the center of the circle is (0,0).
Plug the center and the radius into the equation:
[tex](x-h)^2+(y-k)^2=r^2\\x^2+y^2=6^2\\x^2+y^2=36[/tex]
Answer
[tex]x^2+y^2=36[/tex]
