Find the power of point (2,4) with respect to the circle `x^(2)+y^(2)-6x+4y-8=0`

Answer

To find the power of the point (2, 4) with respect to the circle given by the equation \(x^2 + y^2 - 6x + 4y - 8 = 0\), we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 6x + 4y - 8 = 0 \]
We can rearrange it as:
\[ x^2 - 6x + y^2 + 4y = 8 \]

### Step 2: Complete the Square
Next, we complete the square for both \(x\) and \(y\):

For \(x\):
\[ x^2 - 6x = (x - 3)^2 - 9 \]

For \(y\):
\[ y^2 + 4y = (y + 2)^2 - 4 \]

Substituting these into the equation gives:
\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 = 8 \]
Simplifying this:
\[ (x - 3)^2 + (y + 2)^2 - 13 = 8 \]
\[ (x - 3)^2 + (y + 2)^2 = 21 \]

### Step 3: Identify the Center and Radius
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center \((h, k) = (3, -2)\)
- Radius \(r = \sqrt{21}\)

### Step 4: Calculate the Power of the Point
The power of a point \((x_1, y_1)\) with respect to a circle is given by the formula:
\[ \text{Power} = (x_1 - h)^2 + (y_1 - k)^2 - r^2 \]
Substituting \( (x_1, y_1) = (2, 4) \), \(h = 3\), \(k = -2\), and \(r^2 = 21\):
\[ \text{Power} = (2 - 3)^2 + (4 - (-2))^2 - 21 \]
Calculating each term:
\[ = (-1)^2 + (4 + 2)^2 - 21 \]
\[ = 1 + 6^2 - 21 \]
\[ = 1 + 36 - 21 \]
\[ = 16 \]

### Final Answer
The power of the point (2, 4) with respect to the circle is \(16\).

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