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The equation of the circle of radius 3 t...

The equation of the circle of radius 3 that lies in the fourth quadrant and touching the lines `x=0 and y=0,` is

A

`x^(2)+y^(2)-6x+6y+9=0`

B

`x^(2)+y^(2)-6x-6y+9=0`

C

`x^(2)+y^(2)+6x-6y+9=0`

D

`x^(2)+y^(2)+6x+6y+9=0`

Text Solution

Verified by Experts

The correct Answer is:
A
Since, the circel is touching both the coordinate axes in fourth quadrant, so equation is
`(x-3)^(2)+(y+3)^(2)=3^(2)`
`impliesx^(2)+y^(2)-6x+6y+9=0`
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