Question

Segments AB and BC are both tangent to the circle shown above. What is the value of x? *Note: photo not necessarily to scale

Answer 1

In right triangle OAB and right triangle OCB above.

$OA=OC$

Therefore,

$\begin{gathered} m\angle OAB=m\angle OCB=90^0 \\ Then, \\ OB=OB \end{gathered}$

Hence,

$\begin{gathered} Rt\Delta OAB\cong Rt\Delta OCB \\ \therefore AB=CB \end{gathered}$

Then,

$\begin{gathered} x^2+3=12 \\ x^2=12-3=9 \\ x^2=9 \\ x=\sqrt{9}=3 \end{gathered}$

Hence,

$x=3$