Given that PT is a tangent to the small circle and PS is a tangent to the big circle DPTB and DPSA are right -angled at T and S respectively.
Question = Find the length of ST.

Statement 1 = The radius of the bigger and smaller circles are 9cm and 4cm respectively .
This makes the distance between AS = 9cm and BT =4cm.
By adding a perpendicular line BD on side AS makes BDST a rectangle and DADB a right-angle triangle.
Since the opposites sides of rectangles are equal, then from BDST we have BT = SD = 4cm
If AS = 9cm and SD = 4cm
AD = 9-4 = 5cm, AB = 9 + 4 = 13cm
In right angle triangle DADB we will calculate the length of BD using Pythagoras theories.
$$h_2^2\ =\ 0^2\ +\ a^2$$
$$\left(AB\right)^2\ =\ \left(BD\right)^2\ +\ \left(AD\right)^2$$
Making BD the subject of formula.
$$BD\sqrt{\left(AD\right)^2\ -\ \left(AD\right)^2}$$
$$BD\sqrt{\left(13\right)^2\ -\ \left(5\right)^2}$$
$$BD\sqrt{144}=\ 12cm$$
Remember that opposite sides in a rectangle are equal so from rectangle BDST.
BT = SD = 4cm
BT = ST = 12cm
Hence, Statement 1 is SUFFICIENT.

$$Statement\ 2\ =\ PB\ =\ \frac{52}{5}cm$$
In right angle triangle BTP we only know the length of PB to find the length of the other sides of this triangle we need to know other
length of the other sides of this triangle, we need to know either one of the unknown angles or unknown sides,
but we were not given enough information to calculate the distance ST, hence Statement 2 is NOT SUFFICIENT.

Option A is CORRECT.

AAPL wrote:Source: e-GMAT



A bigger circle (with center A) and a smaller (circle with center B) are touching each other externally. PT and PS are tangents drawn to these circles from an external point (as shown in the figure). What is the length of ST?

(1) The radii of the bigger and the smaller circles are 9cm and 4cm respectively.
(2) PB = 52/5 cm.

\[? = ST\]

(1) Sufficient:


\[\Delta PTB\,\, \cong \,\,\,\Delta PSA\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \begin{gathered}
\,\frac{4}{9} = \,\frac{{4 + {\text{aux}}}}{{9 + 4 + 4 + {\text{aux}}}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{aux}}\,\,\,{\text{unique}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,PT\,\,\,{\text{unique}} \hfill \\
\,\frac{9}{4} = \frac{{ST + PT}}{{PT}}\,\,\,\,\,\mathop \Rightarrow \limits^{PT\,\,{\text{unique}}} \,\,\,\,?\,\, = \,\,ST\,\,{\text{unique}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{SUFF}}. \hfill \\
\end{gathered} \right.\]
\[\left( * \right)\,\,\,\Delta PTB\,\,\,\left\{ \begin{gathered}
TB = 4 \hfill \\
\left( {{\text{4}}\,{\text{ + }}\,{\text{aux}}} \right)\,\,{\text{unique}} \hfill \\
\end{gathered} \right.\,\,\,\,\mathop \Rightarrow \limits^{{\text{Pythagoras}}} \,\,\,\,\,PT\,\,\,\,{\text{unique}}\]

(2) Insufficient:




This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.