Find the equation to that chord of the circle ` x^(2) + y^(2) = 81 ` which is bisected at the point (-2,3) and its pole with respect to the circle

Answer

To solve the problem of finding the equation of the chord of the circle \( x^2 + y^2 = 81 \) which is bisected at the point (-2, 3) and its pole with respect to the circle, we will follow these steps:

### Step 1: Write the equation of the circle
The given circle can be expressed in the standard form:
\[ x^2 + y^2 - 81 = 0 \]
Let \( S = x^2 + y^2 - 81 \).

### Step 2: Identify the midpoint of the chord
The midpoint of the chord is given as the point (-2, 3). This means that the coordinates \( (x_1, y_1) \) are:
\[ x_1 = -2, \quad y_1 = 3 \]

### Step 3: Use the chord equation formula
The equation of the chord that is bisected at the point \( (x_1, y_1) \) can be derived using the formula:
\[ T = S_1 \]
where \( T \) is given by:
\[ T: x \cdot x_1 + y \cdot y_1 - 81 = 0 \]
Substituting \( x_1 \) and \( y_1 \):
\[ T: x \cdot (-2) + y \cdot 3 - 81 = 0 \]
This simplifies to:
\[ -2x + 3y - 81 = 0 \]

### Step 4: Calculate \( S_1 \)
Next, we need to find \( S_1 \) by substituting \( x_1 \) and \( y_1 \) into \( S \):
\[ S_1 = (-2)^2 + (3)^2 - 81 = 4 + 9 - 81 = -68 \]

### Step 5: Set up the equation \( T = S_1 \)
Now we can set up the equation:
\[ -2x + 3y - 81 = -68 \]
This simplifies to:
\[ -2x + 3y - 13 = 0 \]
Multiplying through by -1 gives:
\[ 2x - 3y + 13 = 0 \]

### Step 6: Find the pole of the chord
To find the pole of the chord with respect to the circle, we will use the relationship between the polar and the chord. The polar of a point \( (α, β) \) with respect to the circle \( x^2 + y^2 = r^2 \) is given by:
\[ T = 0 \quad \text{where} \quad T: xα + yβ - r^2 = 0 \]
Here \( r^2 = 81 \).

### Step 7: Set up the equations for the pole
From the chord equation \( 2x - 3y + 13 = 0 \), we can compare coefficients:
1. Coefficient of \( x \): \( 2 = \frac{81}{α} \)
2. Coefficient of \( y \): \( -3 = \frac{81}{β} \)
3. Constant term: \( 13 = -81 \)

### Step 8: Solve for \( α \) and \( β \)
From \( 2 = \frac{81}{α} \):
\[ α = \frac{81}{2} \]
From \( -3 = \frac{81}{β} \):
\[ β = -\frac{81}{3} = -27 \]

### Step 9: Write the final answer
Thus, the pole of the chord with respect to the circle is:
\[ \left( \frac{81}{2}, -27 \right) \]

### Summary of the solution:
1. The equation of the chord bisected at (-2, 3) is \( 2x - 3y + 13 = 0 \).
2. The pole of the chord with respect to the circle is \( \left( \frac{81}{2}, -27 \right) \).