The equation of the circle of radius 3 that l | vedclass

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Question

(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 $(h, k

Vedclass · Accepted Answer

$x^2+y^2-6x+6y+9=0$

Vedclass · Answer

(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 $(h, k) = (3, -3)$. The standard equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$. Substituting the values,we get $(x-3)^2 + (y-(-3))^2 = 3^2$. $(x-3)^2 + (y+3)^2 = 9$. Expanding the terms: $(x^2 - 6x + 9) + (y^2 + 6y + 9) = 9$. $x^2 + y^2 - 6x + 6y + 18 = 9$. $x^2 + y^2 - 6x + 6y + 9 = 0$.

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

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