SOLUTION: Find the equation, in standard form, of the circle. Diameter with endpoints (-2,-1) and (6,-3)

Question 1183676: Find the equation, in standard form, of the circle.
Diameter with endpoints (-2,-1) and (6,-3)
Found 3 solutions by josgarithmetic, ikleyn, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
Diameter with endpoints (p,v) and (q,w)
--------
--------

Center point for the circle is some (h,k) and , and .

Radius squared is .

Equation would be .
Using the actual values as you were given may be a little easier.

Your circle's center is ( (-2+6)/2, (-1-3)/2 )
or (2,-2).
.
.

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
Find the equation, in standard form, of the circle.
Diameter with endpoints (-2,-1) and (6,-3)
~~~~~~~~~~~~~~~~~

The center's x-coordinate is = = 2. The center's y-coordinate is = = -2. The square of the diameter's length is d^2 = ((6-(-2))^2 + ((-3)-(-1))^2 = (8^2 + 2^2) = 68. The square of the radius is r^2 = 68/2^2 = 68/4 = 17. The standard form equation of the circle is + = 17.

Solved.

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The post by @josgarithmetic is TOTALLY WRONG.

IGNORE it, for the sake of your safety.

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Find the equation, in standard form, of the circle.
Diameter with endpoints (-2,-1) and (6,-3)

If you want to confuse someone or yourself to death, then use the many variables that the other person presented to you.
If not, then just use the MIDPOINT formula: , to get:
<===== CENTER coordinates of circle
Standard form of the equation of a CIRCLE:
------ In the ABOVE, substituting (2, - 2) for (h, k) and (- 2, - 1) for (x, y). NOTE that (6, - 3) could also be used for (x, y)

For the equation in Standard form, we finally get:
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