Answer:
Step-by-step explanation:
The tangent-secant theorem states that if the segments of a secant segment and a tangent segment share an endpoint outside of the circle, the product of the lengths of the secant segment and its external segment equals the square of the length of the tangent segment.
We can go ahead and solve for x as shown below;
![\begin{gathered} (x-3)16=(x+3)^2 \\ 16x-48=x^2+6x+9 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/gio38is2mnvxd91imd5f4uzsror0f1zsa0.png)
![\begin{gathered} x^2-16x+6x+9+48=0 \\ x^2-10x+57=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/r7g7twbaizrg2wq4mv1q2kn84n1tqms2y0.png)