Locate the position of the point P with
resect to the circle `S = 0 ` when
`P(4,2) and S-=2x^(2)+2y^(2)-5x-4y-3=0`

Answer

To locate the position of the point \( P(4, 2) \) with respect to the circle defined by the equation \( S = 2x^2 + 2y^2 - 5x - 4y - 3 = 0 \), we will follow these steps:

### Step 1: Understand the Circle's Equation
The equation of the circle is given as:
\[ S = 2x^2 + 2y^2 - 5x - 4y - 3 = 0 \]
We need to evaluate this equation at the point \( P(4, 2) \).

### Step 2: Substitute the Coordinates of Point P into the Circle's Equation
Substituting \( x = 4 \) and \( y = 2 \) into the equation:
\[ S(4, 2) = 2(4^2) + 2(2^2) - 5(4) - 4(2) - 3 \]

### Step 3: Calculate Each Term
Calculating each term step-by-step:
1. \( 4^2 = 16 \) so \( 2(4^2) = 2 \times 16 = 32 \)
2. \( 2^2 = 4 \) so \( 2(2^2) = 2 \times 4 = 8 \)
3. \( -5(4) = -20 \)
4. \( -4(2) = -8 \)

### Step 4: Combine All the Terms
Now, combine all the calculated terms:
\[ S(4, 2) = 32 + 8 - 20 - 8 - 3 \]
Calculating this step-by-step:
- \( 32 + 8 = 40 \)
- \( 40 - 20 = 20 \)
- \( 20 - 8 = 12 \)
- \( 12 - 3 = 9 \)

### Step 5: Analyze the Result
We find that:
\[ S(4, 2) = 9 \]
Since \( S(4, 2) > 0 \), this indicates that the point \( P(4, 2) \) lies outside the circle.

### Conclusion
The position of the point \( P(4, 2) \) with respect to the circle is that it lies **outside** the circle.

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