A) \[10\]
B) 15
C) 5
D) none of these
Correct Answer: B
Solution :
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 \[PC=\sqrt{{{(10-2)}^{2}}+{{(7-1)}^{2}}}\] \[=10\] \[BC=\sqrt{4+1+20}\] \[=5\] \[PB=PC+CB\] \[=10+5\] \[=15\]You need to login to perform this action.
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