question_answer

The greatest distance of the point  from the circle \[{{x}^{2}}+{{y}^{2}}-4x-2y-20=0\]is

A)  \[10\]           

B)  \[15\]

C)   \[5\]            

D)  None of these

Correct Answer: B

Solution :

Given circle is \[{{x}^{2}}+{{y}^{2}}-4x-2y-20=0\] \[\therefore \] Centre is \[C(2,1)\] and radius, \[r=\sqrt{{{2}^{2}}+{{1}^{2}}-(-20)}=5\] At point \[P\,\,(10,7),\] \[{{S}_{1}}\equiv {{10}^{2}}+{{7}^{2}}-40-14-20>0\] \[\therefore \]  The point lies outside the circle. Now, \[CP=\sqrt{{{(10-2)}^{2}}+{{(7-1)}^{2}}}\] \[=\sqrt{64+36}=10\] \[\therefore \] The greatest distance, \[AP=AC+CP=5+10=15\]