In a circle with centre ' $O^{'}PA$ and $PB$ are two chords. $PC$ is the chord that bisects the

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In a circle with centre ' $O^{'}PA$ and $PB$ are two chords. $PC$ is the chord that bisects the angle $APB$. The tangent to the circle at $C$ is drawn meeting $PA$ and $PB$ extended at $Q$ and $R$ respectively. If $QC = 3,QA = 2$ and $RC = 4$, then length of $RB$ equals:

Accepted Answer

[Key Concept: Use the bisected-chord property: the line from the centre to the midpoint is perpendicular to the chord.]$(PQ)(AQ) = (QC)^{2}$$$\Rightarrow \ PQ = \frac{9}{2}$$$$\begin{matrix} \frac{QC}{RC} & \ = \frac{PQ}{PR} \Rightarrow PR = 6 \ (RC)^{2} & \ = (RB)(RP) \Rightarrow RB = \frac{8}{3} \end{matrix}$$

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