Locate the position of the point P with respect to the circle S=0 when
(i) P(1,2) and `S=x^(2)+y^(2)+6x+8y-96`
(ii) P(3,4) and `S=x^(2)+y^(2)-4x-6y-12`
(iii) P(2,-1) and `S=x^(2)+y^(2)-2x-4y+3`
(iv) P(1,5) and `S=x^(2)+y^(2)-2x-4y+3`

Answer

To locate the position of the point \( P \) with respect to the circle defined by the equation \( S = 0 \), we will substitute the coordinates of the point \( P(x, y) \) into the equation of the circle and analyze the result.

### Step-by-Step Solution:

**(i) For \( P(1, 2) \) and \( S = x^2 + y^2 + 6x + 8y - 96 \)**

1. Substitute \( P(1, 2) \) into the equation:
\[ S = 1^2 + 2^2 + 6(1) + 8(2) - 96 \]
2. Calculate each term:
\[ S = 1 + 4 + 6 + 16 - 96 \]
\[ S = 27 - 96 = -69 \]
3. Since \( S < 0 \), the point \( P(1, 2) \) lies **inside the circle**.

---

**(ii) For \( P(3, 4) \) and \( S = x^2 + y^2 - 4x - 6y - 12 \)**

1. Substitute \( P(3, 4) \) into the equation:
\[ S = 3^2 + 4^2 - 4(3) - 6(4) - 12 \]
2. Calculate each term:
\[ S = 9 + 16 - 12 - 24 - 12 \]
\[ S = 25 - 48 = -23 \]
3. Since \( S < 0 \), the point \( P(3, 4) \) lies **inside the circle**.

---

**(iii) For \( P(2, -1) \) and \( S = x^2 + y^2 - 2x - 4y + 3 \)**

1. Substitute \( P(2, -1) \) into the equation:
\[ S = 2^2 + (-1)^2 - 2(2) - 4(-1) + 3 \]
2. Calculate each term:
\[ S = 4 + 1 - 4 + 4 + 3 \]
\[ S = 8 \]
3. Since \( S > 0 \), the point \( P(2, -1) \) lies **outside the circle**.

---

**(iv) For \( P(1, 5) \) and \( S = x^2 + y^2 - 2x - 4y + 3 \)**

1. Substitute \( P(1, 5) \) into the equation:
\[ S = 1^2 + 5^2 - 2(1) - 4(5) + 3 \]
2. Calculate each term:
\[ S = 1 + 25 - 2 - 20 + 3 \]
\[ S = 7 \]
3. Since \( S > 0 \), the point \( P(1, 5) \) lies **outside the circle**.

---

### Summary of Results:
- \( P(1, 2) \) lies **inside** the circle.
- \( P(3, 4) \) lies **inside** the circle.
- \( P(2, -1) \) lies **outside** the circle.
- \( P(1, 5) \) lies **outside** the circle.