Question

The equation of the circle of radius $3$ that lies in the fourth quadrant and touches 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. $x^2+y^2+6x+6y+9=0$

Answer 1

The correct option is A

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

Explanation for the correct option:

Find the equation of the circle:

Given,

The circle touches the lines $x=0$ and $y=0$ and

The radius of the circle is $3$.

The circle lies in the fourth quadrant.

Which implies,

The circle's centre will has a positive x-coordinate and a negative y-coordinate.

Since the radius of the circle is $3$, then that the centre of the circle will be $(3,-3)$

The general form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Here, $h=3,k=-3$ and $r=3$.

Substitute the values in the general form

We get,

The equation of the circle will be:

$(x-{h)}^2+(y-{k)}^2=r^2$

$$\begin{gathered} (x-{3)}^2+(y-{(-3))}^2=3^2 \\ \Rightarrow {(x-3)}^2+{(y+3)}^2=9 \\ \Rightarrow x^2+y^2-6x+6y+9=0 \end{gathered}$$

Hence, option (A) is the correct answer.