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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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Knowledge Check

  • The radius of the larger circle lying in the first quadrant and touching the line 4x+3y-12=0 and the coordinate axes, is

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    B
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    `(x-5)^(2)+(y+5)^(2)=25`
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    `(x+4)^(2)+(y+4)^(2)=25`
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    `(x+6)^(2)+(y+6)^(2)=25`
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    `x^(2)+y^(2)-8x-8y+24=0`
    B
    `x^(2)+y^(2)=8`
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