The line from the center perpendicular to the chord ST, bisects it.
ST = 10 ( Given)
Therefore,
[tex]TV=\frac{1}{2}\times ST=\frac{1}{2}\times10=5[/tex]
Let the radius be r., then
[tex]r^2=TV^2+VZ^2=5^2+16^2=25+256=281[/tex]
Also,
[tex]\begin{gathered} UZ^2+5^2=r^2 \\ \text{then} \\ (3x+4)^2+25=281 \\ \text{ this implies that} \\ (3x+4)^2=281-25=256 \\ \text{ Therefore} \\ 3x+4=\sqrt[]{256} \end{gathered}[/tex][tex]\begin{gathered} 3x+4=16 \\ \text{then} \\ 3x=16-4=12 \\ \text{therefore} \\ x=4 \end{gathered}[/tex]
