A circle has radius 3 units and its centre lies on the line y=x-1. Find the equation of the circle if it passes through (7,3).

Answer

To find the equation of the circle with a radius of 3 units, whose center lies on the line \( y = x - 1 \) and passes through the point \( (7, 3) \), we can follow these steps:

### Step 1: Define the center of the circle
Let the center of the circle be \( (x_1, y_1) \). Since the center lies on the line \( y = x - 1 \), we can express \( y_1 \) in terms of \( x_1 \):
\[ y_1 = x_1 - 1 \]
Thus, the center can be represented as \( (x_1, x_1 - 1) \).

### Step 2: Use the distance formula
The distance from the center of the circle to the point \( (7, 3) \) must equal the radius of the circle, which is 3 units. Therefore, we can set up the equation:
\[ \sqrt{(x_1 - 7)^2 + (y_1 - 3)^2} = 3 \]

### Step 3: Substitute \( y_1 \) into the distance equation
Substituting \( y_1 = x_1 - 1 \) into the distance equation gives:
\[ \sqrt{(x_1 - 7)^2 + ((x_1 - 1) - 3)^2} = 3 \]
This simplifies to:
\[ \sqrt{(x_1 - 7)^2 + (x_1 - 4)^2} = 3 \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides results in:
\[ (x_1 - 7)^2 + (x_1 - 4)^2 = 9 \]

### Step 5: Expand both squares
Expanding the equation gives:
\[ (x_1^2 - 14x_1 + 49) + (x_1^2 - 8x_1 + 16) = 9 \]
Combining like terms results in:
\[ 2x_1^2 - 22x_1 + 65 = 9 \]

### Step 6: Rearrange the equation
Rearranging the equation gives:
\[ 2x_1^2 - 22x_1 + 56 = 0 \]

### Step 7: Simplify the quadratic equation
Dividing the entire equation by 2 simplifies it to:
\[ x_1^2 - 11x_1 + 28 = 0 \]

### Step 8: Factor the quadratic equation
Factoring the quadratic gives:
\[ (x_1 - 7)(x_1 - 4) = 0 \]
Thus, we find:
\[ x_1 = 7 \quad \text{or} \quad x_1 = 4 \]

### Step 9: Find corresponding \( y_1 \) values
Using \( y_1 = x_1 - 1 \):
- If \( x_1 = 7 \), then \( y_1 = 6 \).
- If \( x_1 = 4 \), then \( y_1 = 3 \).

### Step 10: Write the equations of the circles
Now, we can write the equations of the circles using the standard form \( (x - x_1)^2 + (y - y_1)^2 = r^2 \):
1. For \( (7, 6) \):
\[ (x - 7)^2 + (y - 6)^2 = 9 \]
2. For \( (4, 3) \):
\[ (x - 4)^2 + (y - 3)^2 = 9 \]

### Final Equations
The equations of the circles are:
1. \( (x - 7)^2 + (y - 6)^2 = 9 \)
2. \( (x - 4)^2 + (y - 3)^2 = 9 \)