In the adjoining figure, $$PA, PB$$ are tangents. $$AR$$ is parallel to $$PB$$.

$$PQ = 6$$; $$QR = 18$$.
Length $$SB$$ = ------.

Using the tangent-secant property, we get

$$ PA^2 = PB^2 = PQ \times PR = 6 \times 24 = 144$$

$$ PA = PB = 12$$cm

Consider $$\triangle PQS, \triangle RQA$$

$$\angle PQS = \angle RQA $$(Vertically opposite angles)

$$\angle SPQ = \angle ARQ$$ (Interior alternate angles)

$$\angle PSQ = \angle RAQ$$( Interior alternate angles)

Thus, by AAA similarity criterion

$$\triangle PQS \sim \triangle RQA$$

$$\dfrac{QS}{QA} = \dfrac{PS}{RA}= \dfrac{PQ}{RQ} =\dfrac{1}{3}$$ -Eq 1

Now consider $$\triangle PAQ, \triangle PRA$$
$$ \angle RPA = \angle APQ$$ (Common angle)

$$\angle PRA = \angle PAQ$$ (Alternate segment theorem)

Using the AA similarity criterion,

$$\triangle PAQ \sim \triangle PRA$$

$$\dfrac{AQ}{RA}= \dfrac{PQ}{PA} =\dfrac{PA}{PR} = \dfrac{1}{2}$$

Let $$ RA =6x, QA =3x$$

Using Eq 1 we get

$$QS = x, PS = 2x$$

Now we can once again apply tangent-secant theorem to find SB

$$ SB = 12 - PS = 12-2x$$

$$SB^2 = SQ\times SA = 4x^2 \implies SB = 2x$$

Equating the two we get

$$12-2x = 2x \implies x=3 ,SB = 2x = 6$$