Question 1183676
.
Find the equation, in standard form, of the circle.
Diameter with endpoints (-2,-1) and (6,-3)
~~~~~~~~~~~~~~~~~


<pre>
The center's  x-coordinate is  {{{(-2 + 6)/2}}} = {{{4/2}}} = 2.

The center's  y-coordinate is  {{{(-1 + (-3))/2}}} = {{{-4/2}}} = -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


    {{{(x-2)^2}}} + {{{(y+2)^2}}} = 17.
</pre>

Solved.



//////////////



The post by @josgarithmetic is &nbsp;TOTALLY &nbsp;WRONG.



IGNORE &nbsp;it, &nbsp;for the sake of your safety.