The standard form for the equation of a circle is :
(x−h)^2+(y−k)^2=r2 ----------- EQ(1)
where handk are the x and y coordinates of the center of the circle and r is the radius.
The center of the circle is the midpoint of the diameter.
So the midpoint of the diameter with endpoints at (−3,-2)and(7,-6) is :
((−3+7)/2,(-2+(-6))/2)=(2,-4)
So the point (2,-4) is the center of the circle.
Now, use the distance formula to find the radius of the circle:
r^2=(−3−(2))^2+(-2-(-4))^2=25+4=29
⇒r=√29
Subtituting h=2, k=-4 and r=√29 into EQ(1) gives :
(x-2)^2+(y+4)^2=29