The diameter of a circle has endpoints P(-12, -4) and Q(6, 12). Write an equation for the circle. Be sure to show and explain all work.

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  (x +3)² +(y -4)² = 145

Step-by-step explanation:

The center of the circle is the midpoint of the given segment PQ. If we call that point A, then ...

  A = (P +Q)/2

  A = ((-12, -4) +(6, 12))/2 = (-12+6, -4+12)/2 = (-6, 8)/2

  A = (-3, 4)

The equation of the circle for some radius r is ...

  (x -(-3))² +(y -4)² = r² . . . . . . where (-3, 4) is the center of the circle

The value of r² can be found by substituting either of the points on the circle. If we use Q, then we have ...

  (6 +3)² +(12 -4)² = r² = 9² +8²

  r² = 81 +64 = 145

Then the equation of the circle is ...

  (x +3)² +(y -4)² = 145

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.