A circle of radius 3 units touches both the axes in the first quadrant. Find its equation.

Answer

To find the equation of a circle of radius 3 units that touches both the axes in the first quadrant, we can follow these steps:

### Step 1: Understand the Position of the Circle
Since the circle touches both the x-axis and y-axis in the first quadrant, the center of the circle must be at a distance equal to the radius from both axes.

### Step 2: Determine the Center of the Circle
Given that the radius \( r = 3 \) units, the center of the circle will be at the point \( (3, 3) \). This is because the center must be 3 units away from both the x-axis and the y-axis.

### Step 3: Use the Standard Equation of a Circle
The standard equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = 3 \), \( k = 3 \), and \( r = 3 \) into the equation, we get:
\[ (x - 3)^2 + (y - 3)^2 = 3^2 \]

### Step 4: Simplify the Equation
Now, simplify the equation:
\[ (x - 3)^2 + (y - 3)^2 = 9 \]

### Step 5: Expand the Equation
Expanding the left side:
\[ (x^2 - 6x + 9) + (y^2 - 6y + 9) = 9 \]
Combining like terms gives:
\[ x^2 + y^2 - 6x - 6y + 18 = 9 \]

### Step 6: Rearranging the Equation
To put it in standard form, we can rearrange it:
\[ x^2 + y^2 - 6x - 6y + 9 = 0 \]

### Final Equation
Thus, the equation of the circle is:
\[ x^2 + y^2 - 6x - 6y + 9 = 0 \]

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