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A variable chord of the circle x^2+y^2=4...

A variable chord of the circle `x^2+y^2=4` is drawn from the point `P(3,5)` meeting the circle at the point `A` and `Bdot` A point `Q` is taken on the chord such that `2P Q=P A+P B` . The locus of `Q` is `x^2+y^2+3x+4y=0` `x^2+y^2=36` `x^2+y^2=16` `x^2+y^2-3x-5y=0`

A

`x^(2)+y^(2)+3x+4y=0`

B

`x^(2)+y^(2)=36`

C

`x^(2)+y^(2)=16`

D

`x^(2)+y^(2)-3x-5y=0`

Text Solution

Verified by Experts

The correct Answer is:
4


`2PQ=PA+PB`
or `PQ-PA=PB-PQ`
or `AQ=QB`
Therefore, Q is the midpoint of AB.
Let Q has coordinates (h,k).
Then the equation of chord AB is given by `T= S_(1)`
or `hx+ky-4=h^(2)+k^(2)-4`
This variable chord passes through the point `P(3,5)`. Therefore,
`3h+5k=h^(2)+k^(2`
or `x^(2)+y^(2)-3x-5y=0`
which is the required locus.
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