The point of which the line `9x + y - 28 = 0` is the chord of contact of the circle `2x^2+2y^2-3x+5y-7=0` is

Options

A. (3, 1)
B. (1, 3)
C. (3, -1)
D. (-3, -1)

Answer

To solve the problem, we need to find the point \((x_1, y_1)\) such that the line \(9x + y - 28 = 0\) is the chord of contact for the given circle \(2x^2 + 2y^2 - 3x + 5y - 7 = 0\).

### Step 1: Rewrite the equation of the circle in standard form
The given circle equation is:
\[ 2x^2 + 2y^2 - 3x + 5y - 7 = 0 \]
Dividing the entire equation by 2 gives:
\[ x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{7}{2} = 0 \]

### Step 2: Identify the coefficients \(g\), \(f\), and \(c\)
From the standard form of the circle \(x^2 + y^2 + 2gx + 2fy + c = 0\), we can identify:
- \(g = -\frac{3}{4}\)
- \(f = \frac{5}{4}\)
- \(c = -\frac{7}{2}\)

### Step 3: Write the equation of the chord of contact
The equation of the chord of contact from the point \((x_1, y_1)\) is given by:
\[ x_1x + y_1y + g(x_1 + x) + f(y_1 + y) + c = 0 \]
Substituting the values of \(g\), \(f\), and \(c\):
\[ x_1x + y_1y - \frac{3}{4}(x_1 + x) + \frac{5}{4}(y_1 + y) - \frac{7}{2} = 0 \]

### Step 4: Rearranging the equation
This can be rearranged to:
\[ \left(4x_1 - 3\right)x + \left(4y_1 + 5\right)y - 14 = 0 \]

### Step 5: Set the coefficients equal to those of the given line
The line given in the question is:
\[ 9x + y - 28 = 0 \]
From this, we can compare coefficients:
1. \(4x_1 - 3 = 9\)
2. \(4y_1 + 5 = 1\)

### Step 6: Solve for \(x_1\) and \(y_1\)
From the first equation:
\[ 4x_1 - 3 = 9 \implies 4x_1 = 12 \implies x_1 = 3 \]

From the second equation:
\[ 4y_1 + 5 = 1 \implies 4y_1 = -4 \implies y_1 = -1 \]

### Conclusion
Thus, the point \((x_1, y_1)\) is:
\[ (3, -1) \]