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Two parallel chords in a circle have lengths 6 and 8. 

The distance between them is 1. 

Find the diameter of the circle.

\(\begin{array}{|rcll|} \hline 3^2+(1+x)^2 &=& R^2 \\ 4^2+x^2&=& R^2 \\ \hline R^2 = 3^2+(1+x)^2 &=& 4^2+x^2 \\ 3^2+(1+x)^2 &=& 4^2+x^2 \\ 9+1+2x+x^2 &=& 16+x^2 \\ 9+1+2x &=& 16 \\ 10+2x &=& 16 \\ 2x &=& 6 \\ \mathbf{x} &=& \mathbf{3} \\\\ R^2 &=& 4^2 + x^2 \\ R^2 &=& 16+3^2 \\ R^2 &=& 25 \\ \mathbf{R} &=& \mathbf{5} \\ \mathbf{2R} &=& \mathbf{10} \\ \hline \end{array}\)

The diameter of the circle is 10

If  we  bisect  both chords  from the  center of the circle  and draw  a radius to both chord ends, we  will have two right triangles.....

Call  the distance  from the center of the  circle to the longer chord  =  d

Call the  distance  from the  center of the  circle to the shorter chord  = d + 1

So....we  have  this   system

4^2  +  d^2  =  r^2

3^2  + (d + 1)^2   = r^2

Set the r's  equal

4^2  + d^2  = 3^2  + (d + 1)^2

16 + d^2 =  9  + d^2  + 2d  + 1

16  =  10 +  2d

6   = 2d

d  = 3

And using either  equation  we  have that

4^2  + d^2   =r^2

16 + 9 = r^2

25  = r^2

r  = 5

So....the diameter   = 10