Find the equation of the circle which passes through the two points (6,4),(8,-4) and has centre on the X-axis

Answer

To find the equation of the circle that passes through the points (6, 4) and (8, -4) with its center on the x-axis, we can follow these steps:

### Step 1: Define the Center of the Circle
Since the center of the circle is on the x-axis, we can denote the center as \( (h, 0) \), where \( h \) is the x-coordinate of the center.

### Step 2: Use the Distance Formula
The radius of the circle is the distance from the center to any point on the circle. We can find the radius using the distance formula for both points.

1. For point \( A(6, 4) \):
\[ r^2 = (h - 6)^2 + (0 - 4)^2 \]
Simplifying this gives:
\[ r^2 = (h - 6)^2 + 16 \quad \text{(Equation 1)} \]

2. For point \( B(8, -4) \):
\[ r^2 = (h - 8)^2 + (0 + 4)^2 \]
Simplifying this gives:
\[ r^2 = (h - 8)^2 + 16 \quad \text{(Equation 2)} \]

### Step 3: Set the Two Equations Equal
Since both expressions equal \( r^2 \), we can set them equal to each other:
\[ (h - 6)^2 + 16 = (h - 8)^2 + 16 \]
Subtracting 16 from both sides:
\[ (h - 6)^2 = (h - 8)^2 \]

### Step 4: Solve the Equation
Now, we can expand both sides:
\[ (h - 6)^2 = h^2 - 12h + 36 \]
\[ (h - 8)^2 = h^2 - 16h + 64 \]
Setting them equal gives:
\[ h^2 - 12h + 36 = h^2 - 16h + 64 \]
Subtracting \( h^2 \) from both sides:
\[ -12h + 36 = -16h + 64 \]
Rearranging gives:
\[ 4h = 28 \]
Thus:
\[ h = 7 \]

### Step 5: Find the Radius
Now that we have \( h = 7 \), we can substitute this back into either Equation 1 or Equation 2 to find \( r^2 \):
Using Equation 1:
\[ r^2 = (7 - 6)^2 + 16 = 1 + 16 = 17 \]

### Step 6: Write the Equation of the Circle
The standard form of the equation of a circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = 7 \), \( k = 0 \), and \( r^2 = 17 \):
\[ (x - 7)^2 + (y - 0)^2 = 17 \]
Expanding this:
\[ (x - 7)^2 + y^2 = 17 \]
\[ x^2 - 14x + 49 + y^2 = 17 \]
Rearranging gives:
\[ x^2 + y^2 - 14x + 32 = 0 \]

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

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