To find the inverse point of (-2, 3) with respect to the circle given by the equation \(x^2 + y^2 - 4x - 6y + 9 = 0\), we will follow these steps:
### Step 1: Rewrite the Circle Equation
First, we need to rewrite the given circle equation in standard form. The equation is:
\[
x^2 + y^2 - 4x - 6y + 9 = 0
\]
We can rearrange this to:
\[
x^2 - 4x + y^2 - 6y + 9 = 0
\]
### Step 2: Complete the Square
Next, we complete the square for both \(x\) and \(y\).
For \(x\):
\[
x^2 - 4x = (x - 2)^2 - 4
\]
For \(y\):
\[
y^2 - 6y = (y - 3)^2 - 9
\]
Substituting these back into the equation gives:
\[
(x - 2)^2 - 4 + (y - 3)^2 - 9 + 9 = 0
\]
This simplifies to:
\[
(x - 2)^2 + (y - 3)^2 - 4 = 0
\]
Thus, we have:
\[
(x - 2)^2 + (y - 3)^2 = 4
\]
### Step 3: Identify the Center and Radius
From the standard form of the circle, we can identify the center \((h, k)\) and radius \(r\):
- Center: \((2, 3)\)
- Radius: \(r = \sqrt{4} = 2\)
### Step 4: Calculate the Distance OP
Let \(O\) be the center of the circle \((2, 3)\) and \(P\) be the point \((-2, 3)\).
The distance \(OP\) is calculated using the distance formula:
\[
OP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Substituting the coordinates:
\[
OP = \sqrt{((-2) - 2)^2 + (3 - 3)^2} = \sqrt{(-4)^2 + 0^2} = \sqrt{16} = 4
\]
### Step 5: Use the Inversion Formula
The inverse point \(Q\) satisfies the condition:
\[
OP \cdot OQ = r^2
\]
We know \(OP = 4\) and \(r = 2\), thus:
\[
4 \cdot OQ = 2^2
\]
\[
4 \cdot OQ = 4
\]
\[
OQ = 1
\]
### Step 6: Find the Coordinates of the Inverse Point Q
Since \(O\) and \(P\) are collinear, and the line \(OP\) is horizontal (as both points have the same \(y\)-coordinate), the coordinates of point \(Q\) will be \((h, k)\) where \(k = 3\) and \(h\) is determined by moving 1 unit away from the center \(O\) towards \(P\).
Since \(P\) is to the left of \(O\), we move left from \(O\):
\[
Q_x = 2 - 1 = 1
\]
Thus, the coordinates of the inverse point \(Q\) are:
\[
Q = (1, 3)
\]
### Final Answer
The inverse point of \((-2, 3)\) with respect to the circle is \((1, 3)\).
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