Acircle has a diameter with end points at 3+18i and 7-6i. which point is also on the circle ?

One point that is also on the circle is 7 + 18i.

Step-by-step explanation:

Data:

End points of diameter at 3+ 18i and 7 – 6i.

Calculations

Centre of circle:

The centre is at the middle of the end points.

(x, y) = ((x₁ + x₂)/2, (y₁ + y₂)/2)

        = ((3+7)/2, (18i – 6 i)/2))

        = (10/2, 12i/2)

        = (5, 6i)

Radius:

The centre is at (5, 6i). A point on the circumference is (3, 18i)

The distance between the two points is given by

r² = (x₂ - x₁)² + (y₂ - y₁)²

   = (3 - 5)² + (18 - 6)²

   = (-2)² + (12)²

   = 4 + 144

   = 148

r   = √148

Equation for circle

The general equation for a circle is

(x – h)²  + (y - k)² = r²     Substitute values for h, k

( x - 5)² + (y - 6)² = 148

Points on circle:

Any point that satisfies the above equation  is a point on the circle.

There are many solutions, but one point that works is 7 + 18i (see graph below).

For example,

(7 - 5)² + (18 - 6)² = 2² + 12² = 4 + 144 = 148

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'.