Equation of circle centred on X-axis, and passing through
(6,4) and (8,-4), is

Options

A. `x^(2) + y^(2) + 14x + 32 = 0 `
B. `x^(2) + y^(2) + 14y + 32 =0 `
C. `x^(2) + y^(2) - 14y + 32 = 0 `
D. `x^(2) + y^(2) - 14x + 32 = 0 `

Answer

To find the equation of a circle centered on the X-axis and passing through the points (6, 4) and (8, -4), we can follow these steps:

### Step 1: General Equation of Circle
The general equation of a circle can be written as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where the center of the circle is at the point (-g, -f).

### Step 2: Center on the X-axis
Since the center of the circle is on the X-axis, it can be expressed as (a, 0). Therefore:
- \( g = -a \)
- \( f = 0 \)

Substituting these values into the general equation gives:
\[ x^2 + y^2 - 2ax + c = 0 \]

### Step 3: Substitute the First Point (6, 4)
Now, we will substitute the first point (6, 4) into the equation:
\[ 6^2 + 4^2 - 2a(6) + c = 0 \]
Calculating this:
\[ 36 + 16 - 12a + c = 0 \]
This simplifies to:
\[ 52 - 12a + c = 0 \quad \text{(Equation 1)} \]

### Step 4: Substitute the Second Point (8, -4)
Next, we substitute the second point (8, -4) into the equation:
\[ 8^2 + (-4)^2 - 2a(8) + c = 0 \]
Calculating this:
\[ 64 + 16 - 16a + c = 0 \]
This simplifies to:
\[ 80 - 16a + c = 0 \quad \text{(Equation 2)} \]

### Step 5: Solve the System of Equations
Now we have two equations:
1. \( 52 - 12a + c = 0 \)
2. \( 80 - 16a + c = 0 \)

We can subtract Equation 1 from Equation 2:
\[ (80 - 16a + c) - (52 - 12a + c) = 0 \]
This simplifies to:
\[ 80 - 52 - 16a + 12a = 0 \]
\[ 28 - 4a = 0 \]
Solving for \( a \):
\[ 4a = 28 \implies a = 7 \]

### Step 6: Substitute Back to Find c
Now we substitute \( a = 7 \) back into Equation 1 to find \( c \):
\[ 52 - 12(7) + c = 0 \]
Calculating this:
\[ 52 - 84 + c = 0 \]
\[ c = 84 - 52 = 32 \]

### Step 7: Write the Final Equation
Now we have \( a = 7 \) and \( c = 32 \). The equation of the circle is:
\[ x^2 + y^2 - 2(7)x + 32 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 14x + 32 = 0 \]

### Final Answer
The equation of the circle is:
\[ \boxed{x^2 + y^2 - 14x + 32 = 0} \]