Question 627051
I'm assuming you want to find the equation of the circle.


Recall that the general equation of a circle is {{{(x-h)^2+(y-k)^2=r^2}}}.



So we need the center (h,k) and the radius squared {{{r^2}}}.



First, let's find the center (h,k).



Since the center is the midpoint of the line segment with endpoints (-1,-6) and (1,6), we need to find the midpoint.



X-Coordinate of Midpoint = {{{(x[1]+x[2])/2 = (-1+1)/2=0/2 = 0}}}



Since the x coordinate of midpoint is {{{0}}}, this means that {{{h=0}}}



Y-Coordinate of Midpoint = {{{(y[1]+y[2])/2 = (-6+6)/2=0/2 = 0}}}



Since the y coordinate of midpoint is {{{0}}}, this means that {{{k=0}}}



So the center is the point (0, 0)



---------------------------------------------------



Now let's find the radius squared



Use the formula {{{r^2=(x-h)^2+(y-k)^2}}}, where (h,k) is the center and (x,y) is an arbitrary point on the circle.



In this case, {{{h=0}}} and {{{k=0}}}. Also, {{{x=-1}}} and {{{y=-6}}}. Plug these values into the equation above and simplify to get:



{{{r^2=(-1-0)^2+(-6-0)^2}}}



{{{r^2=(-1)^2+(-6)^2}}}



{{{r^2=1+36}}}



{{{r^2=37}}}



So because  {{{h=0}}}, {{{k=0}}}, and {{{r^2=37}}}, this means that the equation of the circle that passes through the points (-1,-6) and (1,6) (which are the endpoints of the diameter) is 



{{{x^2+y^2=37}}}.


<font color="red">--------------------------------------------------------------------------------------------------------------</font>
If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=I%20Need%20Algebra%20Help">jim_thompson5910@hotmail.com</a>


Also, please consider visiting my website: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a> and making a donation. Thank you


Jim
<font color="red">--------------------------------------------------------------------------------------------------------------</font>