Find the equation of the circle passing through the point `(7,3)` having radius 3 units and whose centre lies on the line `y=x-1`

Answer

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

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

### Step 2: Write the equation of the 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 \( k = h - 1 \) and \( r = 3 \):
\[ (x - h)^2 + (y - (h - 1))^2 = 3^2 \]
This simplifies to:
\[ (x - h)^2 + (y - h + 1)^2 = 9 \]

### Step 3: Substitute the point (7, 3)
Since the circle passes through the point (7, 3), we substitute \( x = 7 \) and \( y = 3 \) into the circle's equation:
\[ (7 - h)^2 + (3 - (h - 1))^2 = 9 \]
This simplifies to:
\[ (7 - h)^2 + (3 - h + 1)^2 = 9 \]
\[ (7 - h)^2 + (4 - h)^2 = 9 \]

### Step 4: Expand and simplify the equation
Expanding both squares:
\[ (7 - h)^2 = 49 - 14h + h^2 \]
\[ (4 - h)^2 = 16 - 8h + h^2 \]
Combining these gives:
\[ 49 - 14h + h^2 + 16 - 8h + h^2 = 9 \]
\[ 2h^2 - 22h + 65 = 9 \]
Subtracting 9 from both sides:
\[ 2h^2 - 22h + 56 = 0 \]

### Step 5: Factor the quadratic equation
Dividing the entire equation by 2:
\[ h^2 - 11h + 28 = 0 \]
Factoring:
\[ (h - 7)(h - 4) = 0 \]
Thus, \( h = 7 \) or \( h = 4 \).

### Step 6: Find corresponding values of \( k \)
Using \( k = h - 1 \):
1. If \( h = 7 \), then \( k = 7 - 1 = 6 \) → Center is \( (7, 6) \).
2. If \( h = 4 \), then \( k = 4 - 1 = 3 \) → Center is \( (4, 3) \).

### Step 7: Write the equations of the circles
1. For center \( (7, 6) \):
\[ (x - 7)^2 + (y - 6)^2 = 9 \]
Expanding:
\[ x^2 - 14x + 49 + y^2 - 12y + 36 = 9 \]
\[ x^2 + y^2 - 14x - 12y + 76 = 0 \]

2. For center \( (4, 3) \):
\[ (x - 4)^2 + (y - 3)^2 = 9 \]
Expanding:
\[ x^2 - 8x + 16 + y^2 - 6y + 9 = 9 \]
\[ x^2 + y^2 - 8x - 6y + 16 = 0 \]

### Final Answer:
The equations of the circles are:
1. \( x^2 + y^2 - 14x - 12y + 76 = 0 \)
2. \( x^2 + y^2 - 8x - 6y + 16 = 0 \)