question_answer

In a circle with centre 'O', PA and PB axe 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

A) \[2\]

B) \[\frac{8}{3}\]

C) \[\frac{10}{3}\] 

D) \[\frac{11}{3}\]

Correct Answer: B

Solution :

\[Q{{C}^{2}}=PQ\times AQ\]

\[PQ=\frac{Q{{C}^{2}}}{AQ}=\frac{9}{2}\]

PC is angle bisector of APQ

\[\therefore \] \[\frac{PQ}{PR}=\frac{QC}{RC}\]

\[PR=\frac{PQ\times RC}{QC}=\frac{\frac{9}{2}\times 4}{3}=6\]

\[R{{C}^{2}}=PR\times BR\]\[\Rightarrow \]\[16=6\times BR\]\[\Rightarrow \]\[BR=\frac{16}{6}=\frac{8}{3}\]