If p and q be the longest distance and the shortest distance respectively of the point (−7,2) from any point (α,β) on the curve whose equation is x2+y2−10x−14y−51=0 then GM of p and q is equal to
A
2√11
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B
5√5
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C
13
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D
none of these
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Solution
The correct option is A2√11 Centre and the radius of the given circle are C(5,7) and r=√25+49+51=5√5 respectively. Let P=(−7,2) be the given point. Now CP=√122+52=13 Thus p=CP+r=13+5√5 and q=CP−r=13−5√5 Hence G.M of p and q is =√pq=√132−125=2√11