Two circles `x^(2)+y^(2)=6` and `x^(2)+y^(2)- 6x+8=0` are given. Then the equation of the circle through their point of intersection and the point (1,1) 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. none of these

Answer

To find the equation of the circle that passes through the points of intersection of the two given circles and the point (1,1), we can follow these steps:

### Step 1: Write the equations of the circles
The first circle is given by:
\[ S_1: x^2 + y^2 = 6 \]

The second circle can be rewritten as:
\[ S_2: x^2 + y^2 - 6x + 8 = 0 \]
This can be rearranged to:
\[ S_2: x^2 + y^2 - 6x + 8 = 0 \]

### Step 2: Combine the equations of the circles
To find the equation of the circle that passes through the intersection points of the two circles, we can use the formula:
\[ S_1 + \lambda S_2 = 0 \]
where \( \lambda \) is a parameter.

Substituting the equations of the circles:
\[ (x^2 + y^2 - 6) + \lambda (x^2 + y^2 - 6x + 8) = 0 \]

### Step 3: Simplify the equation
Expanding the equation:
\[ x^2 + y^2 - 6 + \lambda (x^2 + y^2 - 6x + 8) = 0 \]
This gives:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 - 6 + \lambda(-6x + 8) = 0 \]

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

### Step 5: Substitute \( \lambda \) back into the equation
Now substituting \( \lambda = 1 \) back into the combined equation:
\[ (1 + 1)x^2 + (1 + 1)y^2 - 6 + 1(-6x + 8) = 0 \]
This simplifies to:
\[ 2x^2 + 2y^2 - 6 - 6x + 8 = 0 \]
\[ 2x^2 + 2y^2 - 6x + 2 = 0 \]

### Step 6: Divide by 2
Dividing the entire equation by 2 gives:
\[ x^2 + y^2 - 3x + 1 = 0 \]

### Final Equation
Thus, the equation of the circle that passes through the points of intersection of the two given circles and the point (1,1) is:
\[ x^2 + y^2 - 3x + 1 = 0 \]