Find the equation of a circle of radius 5 units whose centre lies on x-axis and passes through the point (2, 3).

Answer

To find the equation of a circle of radius 5 units whose center lies on the x-axis and passes through the point (2, 3), we can follow these steps:

### Step 1: Understand the General Equation of a Circle
The general equation of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center of the circle and \(r\) is the radius.

### Step 2: Identify the Center of the Circle
Since the center lies on the x-axis, the y-coordinate \(k\) will be 0. Therefore, we can represent the center as:
\[ (h, 0) \]

### Step 3: Substitute the Radius
We know the radius \(r\) is 5 units, so:
\[ r^2 = 5^2 = 25 \]

### Step 4: Substitute the Point into the Circle Equation
The circle passes through the point (2, 3). We substitute \(x = 2\) and \(y = 3\) into the circle's equation:
\[ (2 - h)^2 + (3 - 0)^2 = 25 \]
This simplifies to:
\[ (2 - h)^2 + 9 = 25 \]

### Step 5: Solve for \(h\)
Subtract 9 from both sides:
\[ (2 - h)^2 = 16 \]
Taking the square root of both sides gives us two equations:
\[ 2 - h = 4 \quad \text{or} \quad 2 - h = -4 \]
Solving these:
1. \(2 - h = 4 \implies h = -2\)
2. \(2 - h = -4 \implies h = 6\)

### Step 6: Find the Equations of the Circles
Now we have two possible values for \(h\):
1. **When \(h = 6\)**:
\[ (x - 6)^2 + (y - 0)^2 = 25 \]
Expanding this:
\[ (x - 6)^2 + y^2 = 25 \]
\[ x^2 - 12x + 36 + y^2 = 25 \]
\[ x^2 + y^2 - 12x + 11 = 0 \]

2. **When \(h = -2\)**:
\[ (x + 2)^2 + (y - 0)^2 = 25 \]
Expanding this:
\[ (x + 2)^2 + y^2 = 25 \]
\[ x^2 + 4x + 4 + y^2 = 25 \]
\[ x^2 + y^2 + 4x - 21 = 0 \]

### Final Result
Thus, the equations of the circles are:
1. \(x^2 + y^2 - 12x + 11 = 0\)
2. \(x^2 + y^2 + 4x - 21 = 0\)