Please explain this question

Line segment EF is parallel to the line segment CD. Both line segments have the same lengths ( 8 units ). They are 1 unit apart from each other.

And finally, these line segments create an inscribed rectangle with side lengths of 8 and 1 unit. A diagonal of this rectangle is the diameter of a circle.

 CD = 8          CF = 1                            

 DF = sqrt ( CD² + CF² )    

In the diagram below, chords \(\overline{AB}\) and \(\overline{CD}\) are perpendicular, and meet at X.
If AX = 3, BX = 4, CX = 6, and DX = 2, then find the diameter of the circle.

\(\text{Let the center of the circle $C(x_c,\ y_c)$ }\)

\(\begin{array}{|rcll|} \hline 4+y_c &=& 3-y_c \\ 2y_c &=& -1 \\ \mathbf{y_c} &=& \mathbf{-\dfrac{1}{2}} \\ \hline \end{array} \begin{array}{|rcll|} \hline 6+x_c &=& 2-x_c \\ 2x_c &=& -4 \\ \mathbf{x_c} &=& \mathbf{-2} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & (x-x_c)^2 + (y-y_c)^2 &=& r^2 \\ P(2,0): & (2-x_c)^2 + (0-y_c)^2 &=& r^2 \quad | \quad x_c = -2,\ y_c =-\dfrac{1}{2} \\ & (2+2)^2 + \left(0+\dfrac{1}{2} \right)^2 &=& r^2 \\ & 4^2 + \dfrac{1}{4} &=& r^2 \\ & 16 + \dfrac{1}{4} &=& r^2 \quad | \quad *4 \\ & 64+1 &=& 4r^2 \\ & 65 &=& 4r^2 \quad | \quad \text{sqrt both sides} \\ & \sqrt{65} &=& 2r \\ \hline \end{array}\)

The diameter of the circle is \(\mathbf{\sqrt{65}}\)

Thanks Heureka, 

here is another coordinate geometry approach.

I just found the midpoint of the horizontal line x=-2 and drew a vertical axis of symmetry through it.

then the midpoint of the vertical points y=-0.5 and drew a horizontal axis of symmetry through it.

Where these axes of symmetry meet is the centre.  (-2,-0.5)

Now just use the distance formula between the centre and any point on the circumference to find the radius.

Then do what you need to do to find the diameter.

That is an excellent solution Dragan,  thanks for explaining it :)

Thank you guys for you help! I appreciate it