Question

In the given figure, O is the centre of the circle, AB and CD are the chords of the circle which intersect at P. If $\text{PA = 4 cm, PB = 6 cm, OB = 7 cm and CD}$ is 1 cm greater than twice of OP, then the difference between the length of $PDandPC$ is

• A.
  3 cm
• B.
  4 cm
• C.
  5 cm
• D.
  6 cm

Answer 1

The correct option is C

5 cm
$Given:PA=4cm,PB=6cm,OB=7cmCD=2OP+1$


$ConstructOE\perp AB.$
$Here,AB=PA+PB=4+6=10cm$

Since, the perpendicular drawn from the centre of a circle to the chord bisects the chord.

$\therefore \therefore EA=EB=\frac{AB}{2}=5cmandPE=EA–PA=5–4=1cmIn\Delta OEB,\text{Using Pythagoras theorem,}O{B}^2=O{E}^2+E{B}^2\Rightarrow 7^2=O{E}^2+5^2\Rightarrow O{E}^2=49–25=24\Rightarrow OE=2\sqrt{6}cmIn\Delta PEO,UsingPythagorastheorem,O{E}^2+P{E}^2=O{P}^2\Rightarrow 24+1=O{P}^2\Rightarrow OP=5cm\therefore CD=2OP+1\left(Given\right)=11cmandPA\times PB=PC\times PD(\because ABandCDarethechords)\Rightarrow 4\times 6=x\times (11–x)(LetPC=x)\Rightarrow x^2–11x+24=0\Rightarrow x=3cmorx=8cm\therefore PC=3cmor8cmNow,CD=PC+PD\Rightarrow PD=8cmor3cm$

Thus, the difference between the length of $PDandPCis(8–3)=5cm.$

Hence, the correct answer is option c .