question_answer

N is the foot of the perpendicular from a point P of a circle with radius 7 cm, on a diameter AB of the circle. If the length of the chord PB is 12 cm, then the distance of the point N from the point B is    [SSC (CGL) 2013]

A) \[12\frac{2}{7}cm\]                   

B) \[3\frac{5}{7}\,\,cm\]

C) \[10\frac{2}{7}\,\,cm\]

D) \[6\frac{5}{7}\,\,cm\]

Correct Answer: C

Solution :

On joining AP,

Now, \[\angle APB=90\]

[angle in semi-circle]

In \[\Delta APB,\]applying Pythagoras theorem

\[A{{B}^{2}}=A{{P}^{2}}+P{{B}^{2}}\]

\[\Rightarrow \]   \[{{(14)}^{2}}=A{{P}^{2}}+{{(12)}^{2}}\]

\[\Rightarrow \]   \[A{{P}^{2}}=52\]

Now, let \[AN=x\,\,cm\]

Then,    \[NO=AO-AN=(7-x)\,\,cm\]

In \[\Delta APN,\]\[A{{P}^{2}}={{x}^{2}}+P{{N}^{2}}\]\[\Rightarrow \]\[P{{N}^{2}}=A{{P}^{2}}-{{x}^{2}}\]

\[\Rightarrow \]   \[P{{N}^{2}}=52-{{x}^{2}}\]                                    ... (i)

In \[\Delta PNO,\]\[P{{O}^{2}}=P{{N}^{2}}+N{{O}^{2}}\]  \[[\because PO=7\,m]\]

\[\Rightarrow \]               \[{{(7)}^{2}}=P{{N}^{2}}+{{(7-x)}^{2}}\]

\[\Rightarrow \]   \[49-{{(7-x)}^{2}}=P{{N}^{2}}\]                   (ii)

\[\Rightarrow \]\[44-(49-{{x}^{2}}-14x)=P{{N}^{2}}\]

From Eqs. (i) and (ii), we get

\[52-{{x}^{2}}=49-(49-14x+{{x}^{2}})\]

\[\Rightarrow \]\[52-{{x}^{2}}=49-49+14x-{{x}^{2}}\]\[\Rightarrow \]\[14x=52\]

\[\therefore \]      \[x=\frac{52}{14}=\frac{26}{7}\,\,cm\]

Length of \[NO=7-\frac{26}{7}\,\,cm=\frac{49-26}{7}=\frac{23}{7}\,\,cm\]

So, length of \[NB=NO+OB=7+\frac{23}{7}\]

\[=\frac{49+23}{7}=\frac{72}{7}=10\frac{23}{7}\,\,cm\]