If centre of a circle lies on the line `2x - 6y + 9 =0` and it cuts the circle `x^(2) + y^(2) = 2` orthogonally then the circle passes through two fixed points

Options

A. `((1)/(2),(3)/(2)),((-2)/(5),(6)/(5))`
B. `(2, 3), (-2, 6)`
C. `((-1)/(2), (3)/(2)),((-2)/(5),(6)/(5))`
D. `(-2, 3)(-2, 6)`

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript while providing a clearer structure.

### Step 1: Understand the Problem
We need to find the center of a circle that lies on the line \(2x - 6y + 9 = 0\) and intersects the circle \(x^2 + y^2 = 2\) orthogonally.

### Step 2: Set Up the Circle Equation
Let the equation of the circle be:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where the center of the circle is \((-g, -f)\).

### Step 3: Condition for the Center
Since the center lies on the line \(2x - 6y + 9 = 0\), we substitute \((-g, -f)\) into the line equation:
\[ 2(-g) - 6(-f) + 9 = 0 \]
This simplifies to:
\[ -2g + 6f + 9 = 0 \quad \Rightarrow \quad 2g - 6f - 9 = 0 \]

### Step 4: Orthogonality Condition
The condition for two circles to intersect orthogonally is given by:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]
For the circle \(x^2 + y^2 = 2\), we can rewrite it as:
\[ x^2 + y^2 + 0x + 0y - 2 = 0 \]
Thus, \(g_2 = 0\), \(f_2 = 0\), and \(c_2 = -2\).

Substituting these values into the orthogonality condition gives:
\[ 2g \cdot 0 + 2f \cdot 0 = c + (-2) \quad \Rightarrow \quad 0 = c - 2 \quad \Rightarrow \quad c = 2 \]

### Step 5: Substitute \(c\) into the Circle Equation
Now, substituting \(c = 2\) back into the circle equation:
\[ x^2 + y^2 + 2gx + 2fy + 2 = 0 \]

### Step 6: Substitute \(2g\) from the Line Condition
From the line condition \(2g = 6f + 9\), we substitute \(2g\) into the circle equation:
\[ x^2 + y^2 + (6f + 9)x + 2fy + 2 = 0 \]
This simplifies to:
\[ x^2 + y^2 + 6fx + 9x + 2fy + 2 = 0 \]

### Step 7: Rearranging the Circle Equation
Rearranging gives:
\[ x^2 + y^2 + 6fx + 2fy + 9x + 2 = 0 \]

### Step 8: Substitute \(y = -3x\)
Since \(f\) can be expressed in terms of \(y\), we can substitute \(y = -3x\) (from the line equation derived earlier) into the circle equation:
\[ x^2 + (-3x)^2 + 9x + 2 = 0 \]
This simplifies to:
\[ x^2 + 9x^2 + 9x + 2 = 0 \quad \Rightarrow \quad 10x^2 + 9x + 2 = 0 \]

### Step 9: Solve the Quadratic Equation
Now, we can solve the quadratic equation \(10x^2 + 9x + 2 = 0\) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 10 \cdot 2}}{2 \cdot 10} \]
Calculating the discriminant:
\[ 9^2 - 80 = 81 - 80 = 1 \]
Thus:
\[ x = \frac{-9 \pm 1}{20} \]
Calculating the roots:
1. \(x_1 = \frac{-8}{20} = -\frac{2}{5}\)
2. \(x_2 = \frac{-10}{20} = -\frac{1}{2}\)

### Step 10: Find Corresponding \(y\) Values
Now, substituting back to find \(y\):
1. For \(x = -\frac{2}{5}\):
\[ y = -3\left(-\frac{2}{5}\right) = \frac{6}{5} \]
2. For \(x = -\frac{1}{2}\):
\[ y = -3\left(-\frac{1}{2}\right) = \frac{3}{2} \]

### Final Points
Thus, the two points through which the circle passes are:
1. \(\left(-\frac{2}{5}, \frac{6}{5}\right)\)
2. \(\left(-\frac{1}{2}, \frac{3}{2}\right)\)