Find the equation of the normal at P of the circle S=0 where P and S are given by
(i) `P(3,-4),S=x^(2)+y^(2)+x+y-24`
(ii) `P(1,3),S=3(x^(2)+y^(2))-19x-29y+76`
(iii) `P(3,5),S=x^(2)+y^(2)-10x-2y+6`
(iv) `P(1,2),S=x^(2)+y^(2)-22x-4y+25`

Answer

To find the equation of the normal at point P on the circle defined by the equation S = 0, we will follow these steps for each part of the question.

### Part (i): Given P(3, -4) and S = x² + y² + x + y - 24 = 0

1. **Identify the Circle Equation**:
The given equation of the circle is:
\[ S = x^2 + y^2 + x + y - 24 = 0 \]

2. **Differentiate the Circle Equation**:
We differentiate the equation with respect to x:
\[ \frac{d}{dx}(x^2 + y^2 + x + y - 24) = 0 \]
This gives:
\[ 2x + 2y \frac{dy}{dx} + 1 + \frac{dy}{dx} = 0 \]

3. **Rearranging the Derivative**:
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ (2y + 1)\frac{dy}{dx} = - (2x + 1) \]
Thus,
\[ \frac{dy}{dx} = \frac{- (2x + 1)}{2y + 1} \]

4. **Substituting Point P(3, -4)**:
Substitute \(x = 3\) and \(y = -4\) into the derivative:
\[ \frac{dy}{dx} = \frac{- (2(3) + 1)}{2(-4) + 1} = \frac{- (6 + 1)}{-8 + 1} = \frac{-7}{-7} = 1 \]
So, the slope of the tangent at point P is \(m_1 = 1\).

5. **Finding the Slope of the Normal**:
The slope of the normal \(m_2\) is given by:
\[ m_2 = -\frac{1}{m_1} = -1 \]

6. **Equation of the Normal Line**:
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
where \(P(3, -4)\) and \(m = -1\):
\[ y + 4 = -1(x - 3) \]
Rearranging gives:
\[ y + 4 = -x + 3 \implies x + y + 1 = 0 \]

### Final Equation of Normal:
\[ \text{The equation of the normal is } x + y + 1 = 0 \]

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