The equation of the circle passing through the point (2, -1) and having two diameters along the pair of lines `2x^(2)+6y^(2)-x+y-7xy-1=0` is

Options

A. `x^(2)+y^(2)+10x+6y-19=0`
B. `x^(2)+y^(2)+10x-6y+19=0`
C. `x^(2)+y^(2)+10x+6y+19=0`
D. none

Answer

To find the equation of the circle that passes through the point (2, -1) and has two diameters along the pair of lines given by the equation \(2x^2 + 6y^2 - x + y - 7xy - 1 = 0\), we will follow these steps:

### Step 1: Identify the coefficients from the given equation
The general form of the equation of a pair of straight lines is:
\[ ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0 \]
From the equation \(2x^2 + 6y^2 - x + y - 7xy - 1 = 0\), we can identify:
- \(a = 2\)
- \(b = 6\)
- \(h = -\frac{7}{2}\)
- \(g = -\frac{1}{2}\)
- \(f = \frac{1}{2}\)
- \(c = -1\)

### Step 2: Calculate the intersection point (center of the circle)
The intersection point (center of the circle) can be calculated using the formulas:
\[ x = \frac{hf - bg}{ab - h^2} \]
\[ y = \frac{gh - af}{ab - h^2} \]

Substituting the values:
\[ x = \frac{-\frac{7}{2} \cdot \frac{1}{2} - 6 \cdot -\frac{1}{2}}{2 \cdot 6 - \left(-\frac{7}{2}\right)^2} \]
\[ y = \frac{-\frac{1}{2} \cdot -\frac{7}{2} - 2 \cdot \frac{1}{2}}{2 \cdot 6 - \left(-\frac{7}{2}\right)^2} \]

Calculating \(ab - h^2\):
\[ ab - h^2 = 2 \cdot 6 - \left(-\frac{7}{2}\right)^2 = 12 - \frac{49}{4} = \frac{48 - 49}{4} = -\frac{1}{4} \]

Now substituting back:
\[ x = \frac{-\frac{7}{4} + 3}{-\frac{1}{4}} = \frac{-\frac{7}{4} + \frac{12}{4}}{-\frac{1}{4}} = \frac{\frac{5}{4}}{-\frac{1}{4}} = -5 \]
\[ y = \frac{\frac{7}{4} - 1}{-\frac{1}{4}} = \frac{\frac{7}{4} - \frac{4}{4}}{-\frac{1}{4}} = \frac{\frac{3}{4}}{-\frac{1}{4}} = -3 \]

Thus, the center of the circle is \((-5, -3)\).

### Step 3: Calculate the radius of the circle
The radius can be calculated using the distance formula between the center \((-5, -3)\) and the point \((2, -1)\):
\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ r = \sqrt{(2 - (-5))^2 + (-1 - (-3))^2} = \sqrt{(2 + 5)^2 + (-1 + 3)^2} = \sqrt{7^2 + 2^2} = \sqrt{49 + 4} = \sqrt{53} \]

### Step 4: Write the equation of the circle
The standard form of the equation of a circle with center \((h, k)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = -5\), \(k = -3\), and \(r^2 = 53\):
\[ (x + 5)^2 + (y + 3)^2 = 53 \]

Expanding this:
\[ x^2 + 10x + 25 + y^2 + 6y + 9 = 53 \]
\[ x^2 + y^2 + 10x + 6y + 34 - 53 = 0 \]
\[ x^2 + y^2 + 10x + 6y - 19 = 0 \]

### Final Equation of the Circle
The equation of the circle is:
\[ x^2 + y^2 + 10x + 6y - 19 = 0 \]