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The end points of the diameter of a circle are (6, 0) and (-6, 0). What is the equation of this circle?

Sagot :

theleangreenbean

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]

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