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, \[AP=AC+CP=5+10=15\]You need to login to perform this action.
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