PQ is a chord of the circle `x^(2)+y^(2)-2x-8=0` whose mid-point is (2, 2). The circle passing through P, Q and (1, 2) is

Options

A. `x^(2)+y^(2)-7x+10y+28=0`
B. `x^(2)+y^(2)-7x-10y+22=0`
C. `x^(2)+y^(2)-7x+10y+22=0`
D. `x^(2)+y^(2)+7x+10y-22=0`

Answer

To solve the problem, we need to find the equation of the circle that passes through the points P, Q (the endpoints of the chord) and the point (1, 2). The midpoint of the chord PQ is given as (2, 2).

### Step-by-Step Solution:

1. **Identify the Circle's Equation**:
The equation of the given circle is:
\[ x^2 + y^2 - 2x - 8 = 0 \]

2. **Find the Value of S1**:
The general form of a circle is \( S = x^2 + y^2 - 2hx - 2ky + c = 0 \). Here, \( h = 1 \), \( k = 0 \), and \( c = -8 \). We can substitute the midpoint (2, 2) into the equation to find \( S1 \):
\[ S1 = 2^2 + 2^2 - 2(2) - 8 = 4 + 4 - 4 - 8 = -4 \]

3. **Using the Midpoint to Find the Chord Equation**:
The equation of the chord with midpoint (2, 2) can be derived using the formula \( T = S1 \):
\[ T = 2x + 2y - (x + 2) - 8 = -4 \]
Simplifying gives:
\[ 2x + 2y - x - 2 - 8 = -4 \implies x + 2y - 10 = 0 \]
Thus, the equation of the chord PQ is:
\[ x + 2y - 10 = 0 \]

4. **Equation of the Circle through P, Q, and (1, 2)**:
The equation of the circle that passes through points P and Q can be expressed as:
\[ S + \lambda L = 0 \]
where \( S \) is the original circle's equation and \( L \) is the line equation we just found.

5. **Substituting the Line Equation**:
Substitute \( L = x + 2y - 10 \) into the equation:
\[ x^2 + y^2 - 2x - 8 + \lambda (x + 2y - 10) = 0 \]

6. **Substituting the Point (1, 2)**:
Substitute the point (1, 2) into the equation:
\[ 1^2 + 2^2 - 2(1) - 8 + \lambda(1 + 2(2) - 10) = 0 \]
This simplifies to:
\[ 1 + 4 - 2 - 8 + \lambda(1 + 4 - 10) = 0 \]
\[ -5 + \lambda(-5) = 0 \implies -5 - 5\lambda = 0 \implies \lambda = -1 \]

7. **Final Circle Equation**:
Substitute \( \lambda = -1 \) back into the equation:
\[ x^2 + y^2 - 2x - 8 - (x + 2y - 10) = 0 \]
This simplifies to:
\[ x^2 + y^2 - 3x - 2y + 2 = 0 \]

8. **Rearranging the Equation**:
Rearranging gives:
\[ x^2 + y^2 - 3x - 2y + 2 = 0 \]

### Final Answer:
The equation of the circle passing through points P, Q, and (1, 2) is:
\[ x^2 + y^2 - 3x - 2y + 2 = 0 \]