Locate the position of the point (2,4) w.r.t the circle `x^(2)+y^(2)-5x-6y+11=0`

Answer

To locate the position of the point \( P(2, 4) \) with respect to the circle defined by the equation \( x^2 + y^2 - 5x - 6y + 11 = 0 \), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the circle equation in standard form. The given equation is:
\[ x^2 + y^2 - 5x - 6y + 11 = 0 \]
We can rearrange it as:
\[ x^2 - 5x + y^2 - 6y + 11 = 0 \]

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

For \( x \):
\[ x^2 - 5x \quad \text{can be rewritten as} \quad (x - \frac{5}{2})^2 - \left(\frac{5}{2}\right)^2 = (x - \frac{5}{2})^2 - \frac{25}{4} \]

For \( y \):
\[ y^2 - 6y \quad \text{can be rewritten as} \quad (y - 3)^2 - 3^2 = (y - 3)^2 - 9 \]

Now substituting these back into the equation:
\[ (x - \frac{5}{2})^2 - \frac{25}{4} + (y - 3)^2 - 9 + 11 = 0 \]
Combining the constants:
\[ (x - \frac{5}{2})^2 + (y - 3)^2 - \frac{25}{4} + 2 = 0 \]
\[ (x - \frac{5}{2})^2 + (y - 3)^2 = \frac{25}{4} - 2 \]
\[ (x - \frac{5}{2})^2 + (y - 3)^2 = \frac{25}{4} - \frac{8}{4} = \frac{17}{4} \]

### Step 3: Identify the Center and Radius
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center \( C\left(\frac{5}{2}, 3\right) \)
- Radius \( r = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2} \)

### Step 4: Substitute the Point into the Circle Equation
Now, we will substitute the point \( P(2, 4) \) into the original circle equation to determine its position relative to the circle.
\[ x = 2, \quad y = 4 \]
Substituting into the equation:
\[ 2^2 + 4^2 - 5(2) - 6(4) + 11 \]
Calculating each term:
\[ = 4 + 16 - 10 - 24 + 11 \]
\[ = 20 - 34 + 11 = 20 - 34 + 11 = -3 \]

### Step 5: Analyze the Result
Since the result is:
\[ -3 < 0 \]
This implies that the point \( P(2, 4) \) lies **inside** the circle.

### Final Conclusion
The point \( (2, 4) \) lies inside the circle defined by the equation \( x^2 + y^2 - 5x - 6y + 11 = 0 \).

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