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

x^{2} + y^{2} - 6 x + 6 y + 9 = 0

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x^{2} + y^{2} - 6 x - 6 y + 9 = 0

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x^{2} + y^{2} + 6 x - 6 y + 9 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} + 6 x + 6 y + 9 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A $x^{2} + y^{2} - 6 x + 6 y + 9 = 0$
Given the circle lies in $4^{t h}$ quadrant, and touches $y = 0$ and $x = 0$ and has radius $3.$
$∴$ Equation of circle is
$(x - 3)^{2} + (y + 3)^{2} = 3^{2}$
$\Rightarrow x^{2} + y^{2} - 6 x + 6 y + 9 = 0.$

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