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P is a point on the circle x^2+y^2=9 Q i...

P is a point on the circle `x^2+y^2=9` Q is a point on the line `7x+y+3=0`. The perpendicular bisector of PQ is `x-y+1=0`. Then the coordinates of P are:

A

(3,0)

B

`((72)/(25),-(21)/(25))`

C

(0,3)

D

`(-(72)/(25),(21)/(25))`

Text Solution

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The correct Answer is:
A, D
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