Find the area of the triangle formed by
the normal at `(3,-4)` to the circle
`x^(2) +y^(2) -22x - 4y + 25 = 0` with the co-ordinate axes.

Answer

To find the area of the triangle formed by the normal at the point (3, -4) to the circle given by the equation \(x^2 + y^2 - 22x - 4y + 25 = 0\) with the coordinate axes, we can follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 22x - 4y + 25 = 0 \]
We can complete the square for both \(x\) and \(y\).

For \(x\):
\[ x^2 - 22x \rightarrow (x - 11)^2 - 121 \]

For \(y\):
\[ y^2 - 4y \rightarrow (y - 2)^2 - 4 \]

Substituting these back into the equation gives:
\[ (x - 11)^2 - 121 + (y - 2)^2 - 4 + 25 = 0 \]
\[ (x - 11)^2 + (y - 2)^2 - 100 = 0 \]
\[ (x - 11)^2 + (y - 2)^2 = 100 \]

This represents a circle with center \((11, 2)\) and radius \(10\).

### Step 2: Find the Normal at the Point (3, -4)
The formula for the normal to the circle at a point \((x_1, y_1)\) is given by:
\[ y - y_1 = \frac{f}{x_1 + g}(x - x_1) \]
where \(f\) and \(g\) are derived from the standard form of the circle. Here, \(f = -2\) and \(g = 11\).

Substituting \(x_1 = 3\) and \(y_1 = -4\):
\[ y + 4 = \frac{-2}{3 - 11}(x - 3) \]
\[ y + 4 = \frac{-2}{-8}(x - 3) \]
\[ y + 4 = \frac{1}{4}(x - 3) \]

### Step 3: Simplify the Normal Equation
Rearranging the equation:
\[ y + 4 = \frac{1}{4}x - \frac{3}{4} \]
\[ y = \frac{1}{4}x - \frac{3}{4} - 4 \]
\[ y = \frac{1}{4}x - \frac{3}{4} - \frac{16}{4} \]
\[ y = \frac{1}{4}x - \frac{19}{4} \]

### Step 4: Find Intercepts with the Axes
To find the x-intercept, set \(y = 0\):
\[ 0 = \frac{1}{4}x - \frac{19}{4} \]
\[ \frac{1}{4}x = \frac{19}{4} \]
\[ x = 19 \]

To find the y-intercept, set \(x = 0\):
\[ y = \frac{1}{4}(0) - \frac{19}{4} \]
\[ y = -\frac{19}{4} \]

### Step 5: Area of the Triangle
The area \(A\) of the triangle formed by the intercepts and the origin is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is the x-intercept \(19\) and the height is the absolute value of the y-intercept \(\frac{19}{4}\):
\[ A = \frac{1}{2} \times 19 \times \frac{19}{4} \]
\[ A = \frac{19 \times 19}{8} = \frac{361}{8} = 45.125 \text{ square units} \]

### Final Answer
The area of the triangle formed by the normal at the point (3, -4) to the circle with the coordinate axes is \( \frac{361}{8} \) square units.