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If the circle x^2+y^2+2x+3y+1=0 cuts x^2...

If the circle `x^2+y^2+2x+3y+1=0` cuts `x^2+y^2+4x+3y+2=0` at `A and B` , then find the equation of the circle on `A B` as diameter.

A

`2x^2+2y^2+2x+6y+1=0`

B

`x^2+y^2+x+3y+3=0`

C

`x^2+y^2+x+6y+1=0`

D

`2x^2+2y^2+x+3y+1=0`

Text Solution

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The correct Answer is:
A
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