The equation of circle passing through `(1,-3)` and the points common to the two circt `x^2 + y^2 - 6x + 8y - 16 = 0, x^2 + y^2 + 4x - 2y - 8 = 0` is

Options

A. `x^(2)+y^(2)-4x+6y+24=0`
B. `2x^(2)+2y^(2)+3x+y-20=0`
C. `3x^(2)+3y^(2)-5x+7y-19=0`
D. none of these

Explanation

To find the equation of the circle that passes through the point (1, -3) and the points common to the two given circles, we can follow these steps:

### Step 1: Write down the equations of the given circles
The equations of the two circles are:
1. \( C_1: x^2 + y^2 - 6x + 8y - 16 = 0 \)
2. \( C_2: x^2 + y^2 + 4x - 2y - 8 = 0 \)

### Step 2: Use the family of circles concept
The equation of a circle that passes through the intersection points of the two circles can be expressed as:
\[ C_1 + \lambda C_2 = 0 \]
where \( \lambda \) is a constant.

### Step 3: Substitute the equations into the family of circles
Substituting the equations of \( C_1 \) and \( C_2 \):
\[ (x^2 + y^2 - 6x + 8y - 16) + \lambda (x^2 + y^2 + 4x - 2y - 8) = 0 \]
This simplifies to:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 + (-6 + 4\lambda)x + (8 - 2\lambda)y + (-16 - 8\lambda) = 0 \]

### Step 4: Substitute the point (1, -3)
Since the circle passes through the point (1, -3), we substitute \( x = 1 \) and \( y = -3 \) into the equation:
\[ (1 + \lambda)(1^2) + (1 + \lambda)(-3^2) + (-6 + 4\lambda)(1) + (8 - 2\lambda)(-3) + (-16 - 8\lambda) = 0 \]
Calculating each term:
- \( (1 + \lambda)(1) = 1 + \lambda \)
- \( (1 + \lambda)(9) = 9 + 9\lambda \)
- \( (-6 + 4\lambda)(1) = -6 + 4\lambda \)
- \( (8 - 2\lambda)(-3) = -24 + 6\lambda \)
- \( (-16 - 8\lambda) = -16 - 8\lambda \)

Combining all these:
\[ 1 + \lambda + 9 + 9\lambda - 6 + 4\lambda - 24 + 6\lambda - 16 - 8\lambda = 0 \]
This simplifies to:
\[ (1 + 9 - 6 - 24 - 16) + (1 + 9 + 4 + 6 - 8)\lambda = 0 \]
\[ -36 + 12\lambda = 0 \]

### Step 5: Solve for \( \lambda \)
From the equation:
\[ 12\lambda = 36 \implies \lambda = 3 \]

### Step 6: Substitute \( \lambda \) back into the family of circles equation
Now substituting \( \lambda = 3 \) back into the family of circles equation:
\[ (1 + 3)x^2 + (1 + 3)y^2 + (-6 + 4 \cdot 3)x + (8 - 2 \cdot 3)y + (-16 - 8 \cdot 3) = 0 \]
This gives:
\[ 4x^2 + 4y^2 + (6)x + (2)y - 40 = 0 \]

### Step 7: Simplify the equation
Dividing the entire equation by 2:
\[ 2x^2 + 2y^2 + 3x + y - 20 = 0 \]

### Final Answer
The equation of the circle is:
\[ 2x^2 + 2y^2 + 3x + y - 20 = 0 \]