The equation of circle passing through the point `(1, 1)` and point of intersection `x^(2) + y^(2) = 6` and `x^(2) + y^(2) -6x + 8 =0`, is

Options

A. `x^(2) + y^(2) -6x +4 =0`
B. `x^(2) + y^(2) - 3x + 1 = 0`
C. `x^(2) = y^(2) - 4y + 2 = 0`
D. `x^(2) + y^(2) - 2y + 2 = 0`

Answer

To find the equation of the circle passing through the point (1, 1) and the point of intersection of the two given circles, we will follow these steps:

### Step 1: Identify the equations of the circles
The equations of the two circles are:
1. \( C_1: x^2 + y^2 = 6 \)
2. \( C_2: x^2 + y^2 - 6x + 8 = 0 \)

### Step 2: Rewrite the second circle's equation
We can rewrite the second circle's equation in a standard form:
\[ C_2: x^2 + y^2 - 6x + 8 = 0 \implies x^2 + y^2 = 6x - 8 \]

### Step 3: Find the point of intersection of the two circles
To find the points of intersection, we can set the two equations equal to each other:
\[ 6 = 6x - 8 \]
Solving for \( x \):
\[ 6x = 14 \implies x = \frac{14}{6} = \frac{7}{3} \]

Now substituting \( x = \frac{7}{3} \) back into the first circle's equation to find \( y \):
\[ \left(\frac{7}{3}\right)^2 + y^2 = 6 \implies \frac{49}{9} + y^2 = 6 \]
\[ y^2 = 6 - \frac{49}{9} = \frac{54}{9} - \frac{49}{9} = \frac{5}{9} \]
Thus, \( y = \pm \frac{\sqrt{5}}{3} \).

The points of intersection are:
\[ \left(\frac{7}{3}, \frac{\sqrt{5}}{3}\right) \quad \text{and} \quad \left(\frac{7}{3}, -\frac{\sqrt{5}}{3}\right) \]

### Step 4: Use the general equation of the circle
The general equation of a circle passing through the points of intersection of two circles can be expressed as:
\[ C_1 + \lambda C_2 = 0 \]
Substituting the equations of the circles:
\[ (x^2 + y^2 - 6) + \lambda (x^2 + y^2 - 6x + 8) = 0 \]
Expanding this:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 - 6\lambda x + (8\lambda - 6) = 0 \]

### Step 5: Substitute the point (1, 1)
Since the circle passes through the point (1, 1), we substitute \( x = 1 \) and \( y = 1 \):
\[ (1 + \lambda)(1^2) + (1 + \lambda)(1^2) - 6\lambda(1) + (8\lambda - 6) = 0 \]
This simplifies to:
\[ (1 + \lambda) + (1 + \lambda) - 6\lambda + (8\lambda - 6) = 0 \]
\[ 2 + 2\lambda - 6\lambda + 8\lambda - 6 = 0 \]
\[ (2\lambda - 6 + 2) = 0 \implies 4\lambda - 4 = 0 \implies \lambda = 1 \]

### Step 6: Substitute \( \lambda \) back into the equation
Now substituting \( \lambda = 1 \) back into the equation:
\[ (1 + 1)x^2 + (1 + 1)y^2 - 6(1)x + (8(1) - 6) = 0 \]
This simplifies to:
\[ 2x^2 + 2y^2 - 6x + 2 = 0 \]
Dividing through by 2 gives:
\[ x^2 + y^2 - 3x + 1 = 0 \]

### Final Equation
Thus, the equation of the circle is:
\[ x^2 + y^2 - 3x + 1 = 0 \]