(A) Given,radius $r = 3$. Since the circle lies in the fourth quadrant and touches both coordinate axes ($x=0$ and $y=0$),its center must be at $(3, -

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(A) Given,radius $r = 3$. Since the circle lies in the fourth quadrant and touches both coordinate axes ($x=0$ and $y=0$),its center must be at $(3, -3)$. The standard equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$. Substituting $h=3$,$k=-3$,and $r=3$: $(x-3)^2 + (y-(-3))^2 = 3^2$ $(x-3)^2 + (y+3)^2 = 9$ $x^2 - 6x + 9 + y^2 + 6y + 9 = 9$ $x^2 + y^2 - 6x + 6y + 9 = 0$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

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

AP EAMCET 2007 All AP EAMCET

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$x^2+y^2-6x+6y+9=0$
B
$x^2+y^2-6x-6y+9=0$
C
$x^2+y^2+6x-6y+9=0$
D
$x^2+y^2+6x+6y+9=0$

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