question_answer

The greatest distance of the point \[P(10,\,\,7)\] from the circle \[{{x}^{2}}+{{y}^{2}}-4x-2y-20=0\] is:

A) \[10\]                                   

B) \[15\]

C) \[5\]                                     

D)  none of these

Correct Answer: A

Solution :

Key Idea: If a point is outside the circle, then the greatest distance is equal to the length of diameter and lowest distance between the circle and a point. Since,\[{{S}_{1}}={{10}^{2}}+{{7}^{2}}-4\times 10-2\times 7-20>0.\] So, \[P\] lies outside the circle. Join \[P\] with the centre \[C\,(2,\,\,1)\] of the given circle. Suppose \[PC\] cuts the circle at \[A\] and\[B\], then \[PB\] is the greatest distance of \[P\] from the circle. Now,     \[PC=\sqrt{{{(10-2)}^{2}}+{{(7-1)}^{2}}}=10\]                 \[BC=\sqrt{4+1+20}=5\]                 \[PB=PC+CB\]                        \[=10+5=15\]