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: Identify the equations of the circles

The first circle is given by:
\[ S_1: x^2 + y^2 = 6 \]

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

### Step 2: Write the general equation of the circle through the intersection points

The equation of the new circle that passes through the intersection points of the two circles can be expressed as:
\[ S_1 + \lambda S_2 = 0 \]
Substituting the equations of \( S_1 \) and \( S_2 \):
\[ (x^2 + y^2 - 6) + \lambda (x^2 + y^2 - 6x + 8) = 0 \]

### Step 3: Simplify the equation

Combining the two equations:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 - 6 + \lambda(-6x) + \lambda(8) = 0 \]
This simplifies to:
\[ (1 + \lambda)(x^2 + y^2) - 6\lambda x + (8 - 6) = 0 \]
or:
\[ (1 + \lambda)(x^2 + y^2) - 6\lambda x + 2\lambda = 0 \]

### Step 4: Substitute the point (1,1)

Since the circle must also pass through the point (1,1), we substitute \( x = 1 \) and \( y = 1 \):
\[ (1 + \lambda)(1^2 + 1^2) - 6\lambda(1) + 2\lambda = 0 \]
This simplifies to:
\[ (1 + \lambda)(2) - 6\lambda + 2\lambda = 0 \]
or:
\[ 2 + 2\lambda - 6\lambda + 2\lambda = 0 \]
which simplifies to:
\[ 2 - 2\lambda = 0 \]

### Step 5: Solve for λ

From the equation:
\[ 2 - 2\lambda = 0 \]
we find:
\[ 2\lambda = 2 \]
\[ \lambda = 1 \]

### Step 6: Substitute λ back into the circle equation

Now substituting \( \lambda = 1 \) back into the equation:
\[ (1 + 1)(x^2 + y^2) - 6(1)x + 2(1) = 0 \]
This gives us:
\[ 2(x^2 + y^2) - 6x + 2 = 0 \]

### Step 7: Simplify the equation

Dividing the entire equation by 2:
\[ x^2 + y^2 - 3x + 1 = 0 \]

### Final Result

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

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