(1, 2) is a point on the circle` x^2 + y^2 + 2x - 6y + 5 = 0` which is orthogonal to `x^2 + y^2 = 5`. The conjugate point of (1, 2) w.r.t the circle `x^2+y^2 = 5` and which lies on the first circle is

Options

A. (7,-1)
B. (9,-2)
C. (-3,4)
D. (0,5)

Answer

To solve the problem step by step, we need to find the conjugate point of (1, 2) with respect to the circle \(x^2 + y^2 = 5\) and ensure that this point lies on the circle defined by \(x^2 + y^2 + 2x - 6y + 5 = 0\).

### Step 1: Identify the circles
The first circle is given by:
\[ x^2 + y^2 + 2x - 6y + 5 = 0 \]
The second circle is:
\[ x^2 + y^2 = 5 \]

### Step 2: Rewrite the first circle's equation
We can rewrite the first circle's equation in standard form by completing the square:
\[ x^2 + 2x + y^2 - 6y + 5 = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x + 1)^2 - 1 + (y - 3)^2 - 9 + 5 = 0 \]
This simplifies to:
\[ (x + 1)^2 + (y - 3)^2 = 5 \]
This means the center of the first circle is \((-1, 3)\) and its radius is \(\sqrt{5}\).

### Step 3: Identify the conjugate point
Let the conjugate point of \((1, 2)\) with respect to the circle \(x^2 + y^2 = 5\) be \((x_1, y_1)\). The formula for the conjugate point is:
\[ x_1 = \frac{r^2 x_2}{x_2^2 + y_2^2} \]
\[ y_1 = \frac{r^2 y_2}{x_2^2 + y_2^2} \]
where \((x_2, y_2) = (1, 2)\) and \(r^2 = 5\).

### Step 4: Calculate \(x_1\) and \(y_1\)
Substituting the values:
\[ x_1 = \frac{5 \cdot 1}{1^2 + 2^2} = \frac{5 \cdot 1}{1 + 4} = \frac{5}{5} = 1 \]
\[ y_1 = \frac{5 \cdot 2}{1^2 + 2^2} = \frac{5 \cdot 2}{1 + 4} = \frac{10}{5} = 2 \]
Thus, the conjugate point is \((1, 2)\).

### Step 5: Substitute into the first circle's equation
Since we need the conjugate point to lie on the first circle, we will use the equation:
\[ x_1^2 + y_1^2 + 2x_1 - 6y_1 + 5 = 0 \]
Substituting \(x_1\) and \(y_1\):
\[ (1)^2 + (2)^2 + 2(1) - 6(2) + 5 = 0 \]
Calculating:
\[ 1 + 4 + 2 - 12 + 5 = 0 \]
This simplifies to:
\[ 0 = 0 \]
Thus, the point \((1, 2)\) lies on the first circle.

### Step 6: Find the other conjugate point
Now we need to find another conjugate point. We can use the same formula:
\[ x_1 = \frac{5 \cdot 1}{1^2 + 2^2} = 1 \]
\[ y_1 = \frac{5 \cdot 2}{1^2 + 2^2} = 2 \]
However, we need to find the point that lies on the first circle.

### Step 7: Solve for \(y_1\) in the first circle's equation
Using the equation of the first circle:
\[ x_1^2 + y_1^2 + 2x_1 - 6y_1 + 5 = 0 \]
Substituting \(x_1 = 5 - 2y_1\):
\[ (5 - 2y_1)^2 + y_1^2 + 2(5 - 2y_1) - 6y_1 + 5 = 0 \]
This will give us a quadratic equation in \(y_1\).

### Step 8: Solve the quadratic equation
After simplifying, we will find two possible values for \(y_1\) and then calculate corresponding \(x_1\) values.

### Final Answer
After solving, we find that the conjugate point of \((1, 2)\) with respect to the circle \(x^2 + y^2 = 5\) that lies on the first circle is \((-3, 4)\).