The equation of the circle with radius 3 units, passing through the point (2, 1) and whose centre lies on the line y + x = 0 can be

Options

A. `(x+2)^(2) + (y-2)^(2) = (3)^(2)`
B. `(x-1)^(2) + (y-1)^(2) = (3)^(2)`
C. `(x+2)^(2) + (y-1)^(2) = (3)^(2)`
D. `(x+1)^(2) + (y-1)^(2) = (3)^(2)`

Answer

To find the equation of the circle with a radius of 3 units, passing through the point (2, 1) and whose center lies on the line \(y + x = 0\), we can follow these steps:

### Step 1: Define the center of the circle
Since the center lies on the line \(y + x = 0\), we can express the coordinates of the center as:
\[ (h, -h) \]
where \(h\) is the x-coordinate of the center.

### Step 2: Use the distance formula
The distance from the center \((h, -h)\) to the point \((2, 1)\) must equal the radius of the circle, which is 3 units. We can use the distance formula:
\[ \sqrt{(h - 2)^2 + (-h - 1)^2} = 3 \]

### Step 3: Square both sides
To eliminate the square root, we square both sides:
\[ (h - 2)^2 + (-h - 1)^2 = 9 \]

### Step 4: Expand the equation
Now, we will expand both squares:
\[ (h - 2)^2 = h^2 - 4h + 4 \]
\[ (-h - 1)^2 = h^2 + 2h + 1 \]
Combining these, we have:
\[ h^2 - 4h + 4 + h^2 + 2h + 1 = 9 \]

### Step 5: Simplify the equation
Combine like terms:
\[ 2h^2 - 2h + 5 = 9 \]
Subtract 9 from both sides:
\[ 2h^2 - 2h - 4 = 0 \]

### Step 6: Divide by 2
To simplify, divide the entire equation by 2:
\[ h^2 - h - 2 = 0 \]

### Step 7: Factor the quadratic equation
Next, we will factor the quadratic:
\[ (h - 2)(h + 1) = 0 \]
Thus, we have two possible solutions for \(h\):
\[ h = 2 \quad \text{or} \quad h = -1 \]

### Step 8: Find the corresponding y-coordinates
1. If \(h = 2\):
\[ y = -h = -2 \quad \Rightarrow \quad \text{Center} = (2, -2) \]

2. If \(h = -1\):
\[ y = -h = 1 \quad \Rightarrow \quad \text{Center} = (-1, 1) \]

### Step 9: Write the equations of the circles
Using the center-radius form of the circle's equation:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius.

1. For center \((2, -2)\):
\[ (x - 2)^2 + (y + 2)^2 = 3^2 \quad \Rightarrow \quad (x - 2)^2 + (y + 2)^2 = 9 \]

2. For center \((-1, 1)\):
\[ (x + 1)^2 + (y - 1)^2 = 3^2 \quad \Rightarrow \quad (x + 1)^2 + (y - 1)^2 = 9 \]

### Final Answer
The equations of the circles are:
1. \((x - 2)^2 + (y + 2)^2 = 9\)
2. \((x + 1)^2 + (y - 1)^2 = 9\)