Find the equation of the tangent at P of the circle S=0 where P and S are given by
(i) `P=(3,4),S=x^(2)+y^(2)-4x-6y+11`
(ii) `P=(-1,1),S=x^(2)+y^(2)-5x+4y-12`
(iii) `P=(-6,-9),S=x^(2)+y^(2)+4x+6y-39`
(iv) `P=(7,-5),S=x^(2)+y^(2)-6x+4y-12`

Answer

To find the equations of the tangents at the given points \( P \) for the circles defined by the equations \( S \), we will follow these steps for each part of the question.

### Step-by-Step Solution

#### (i) For \( P = (3, 4) \) and \( S = x^2 + y^2 - 4x - 6y + 11 = 0 \)

1. **Identify the center and radius of the circle**:
- The equation of the circle can be rewritten in standard form:
\[ S = (x^2 - 4x) + (y^2 - 6y) + 11 = 0 \]
Completing the square:
\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 + 11 = 0 \implies (x - 2)^2 + (y - 3)^2 = 2 \]
- Center \( C = (2, 3) \) and radius \( r = \sqrt{2} \).

2. **Find the slope of the radius \( CP \)**:
- Slope \( m_{CP} = \frac{y_P - y_C}{x_P - x_C} = \frac{4 - 3}{3 - 2} = 1 \).

3. **Find the slope of the tangent line**:
- The slope of the tangent line \( m_t \) is the negative reciprocal of \( m_{CP} \):
\[ m_t = -1 \]

4. **Use point-slope form to find the equation of the tangent**:
- Using point \( P(3, 4) \):
\[ y - 4 = -1(x - 3) \implies y - 4 = -x + 3 \implies y = -x + 7 \]

#### (ii) For \( P = (-1, 1) \) and \( S = x^2 + y^2 - 5x + 4y - 12 = 0 \)

1. **Identify the center and radius of the circle**:
- Rewriting:
\[ (x^2 - 5x) + (y^2 + 4y) - 12 = 0 \]
Completing the square:
\[ (x - \frac{5}{2})^2 - \frac{25}{4} + (y + 2)^2 - 4 - 12 = 0 \implies (x - \frac{5}{2})^2 + (y + 2)^2 = \frac{61}{4} \]
- Center \( C = \left(\frac{5}{2}, -2\right) \) and radius \( r = \frac{\sqrt{61}}{2} \).

2. **Find the slope of the radius \( CP \)**:
- Slope \( m_{CP} = \frac{1 + 2}{-1 - \frac{5}{2}} = \frac{3}{-\frac{7}{2}} = -\frac{6}{7} \).

3. **Find the slope of the tangent line**:
- \( m_t = \frac{7}{6} \).

4. **Use point-slope form**:
- Using point \( P(-1, 1) \):
\[ y - 1 = \frac{7}{6}(x + 1) \implies y = \frac{7}{6}x + \frac{7}{6} + 1 = \frac{7}{6}x + \frac{13}{6} \]

#### (iii) For \( P = (-6, -9) \) and \( S = x^2 + y^2 + 4x + 6y - 39 = 0 \)

1. **Identify the center and radius of the circle**:
- Rewriting:
\[ (x^2 + 4x) + (y^2 + 6y) - 39 = 0 \]
Completing the square:
\[ (x + 2)^2 - 4 + (y + 3)^2 - 9 - 39 = 0 \implies (x + 2)^2 + (y + 3)^2 = 52 \]
- Center \( C = (-2, -3) \) and radius \( r = \sqrt{52} = 2\sqrt{13} \).

2. **Find the slope of the radius \( CP \)**:
- Slope \( m_{CP} = \frac{-9 + 3}{-6 + 2} = \frac{-6}{-4} = \frac{3}{2} \).

3. **Find the slope of the tangent line**:
- \( m_t = -\frac{2}{3} \).

4. **Use point-slope form**:
- Using point \( P(-6, -9) \):
\[ y + 9 = -\frac{2}{3}(x + 6) \implies y = -\frac{2}{3}x - 4 - 9 = -\frac{2}{3}x - 13 \]

#### (iv) For \( P = (7, -5) \) and \( S = x^2 + y^2 - 6x + 4y - 12 = 0 \)

1. **Identify the center and radius of the circle**:
- Rewriting:
\[ (x^2 - 6x) + (y^2 + 4y) - 12 = 0 \]
Completing the square:
\[ (x - 3)^2 - 9 + (y + 2)^2 - 4 - 12 = 0 \implies (x - 3)^2 + (y + 2)^2 = 25 \]
- Center \( C = (3, -2) \) and radius \( r = 5 \).

2. **Find the slope of the radius \( CP \)**:
- Slope \( m_{CP} = \frac{-5 + 2}{7 - 3} = \frac{-3}{4} \).

3. **Find the slope of the tangent line**:
- \( m_t = \frac{4}{3} \).

4. **Use point-slope form**:
- Using point \( P(7, -5) \):
\[ y + 5 = \frac{4}{3}(x - 7) \implies y = \frac{4}{3}x - \frac{28}{3} - 5 = \frac{4}{3}x - \frac{43}{3} \]

### Final Results

1. **Tangent equation at \( P(3, 4) \)**: \( y = -x + 7 \)
2. **Tangent equation at \( P(-1, 1) \)**: \( y = \frac{7}{6}x + \frac{13}{6} \)
3. **Tangent equation at \( P(-6, -9) \)**: \( y = -\frac{2}{3}x - 13 \)
4. **Tangent equation at \( P(7, -5) \)**: \( y = \frac{4}{3}x - \frac{43}{3} \)