The equation of the normal at the point (4,-1) of the circle `x^2+y^2-40x+10y=153` is

Options

A. x+4y=0
B. 4x+y=3
C. x-4y=0
D. 4x-y=0

Answer

To find the equation of the normal at the point (4, -1) of the circle given by the equation \(x^2 + y^2 - 40x + 10y = 153\), we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the equation of the circle in standard form. We start with:
\[ x^2 + y^2 - 40x + 10y - 153 = 0 \]
We can complete the square for the \(x\) and \(y\) terms.

### Step 2: Completing the Square
For \(x\):
\[ x^2 - 40x \rightarrow (x - 20)^2 - 400 \]
For \(y\):
\[ y^2 + 10y \rightarrow (y + 5)^2 - 25 \]
Now substituting back into the equation:
\[ (x - 20)^2 - 400 + (y + 5)^2 - 25 - 153 = 0 \]
Simplifying this gives:
\[ (x - 20)^2 + (y + 5)^2 - 578 = 0 \]
Thus, the standard form of the circle is:
\[ (x - 20)^2 + (y + 5)^2 = 578 \]
This indicates that the center of the circle is at (20, -5) and the radius is \(\sqrt{578}\).

### Step 3: Find the Slope of the Radius at (4, -1)
The slope of the radius to the point (4, -1) can be calculated as:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - (-5)}{4 - 20} = \frac{4}{-16} = -\frac{1}{4} \]

### Step 4: Find the Slope of the Normal
The slope of the normal is the negative reciprocal of the slope of the radius:
\[ \text{slope of normal} = -\left(-\frac{1}{4}\right)^{-1} = 4 \]

### Step 5: Use the Point-Slope Form to Find the Normal Equation
Using the point-slope form of the line equation, \(y - y_1 = m(x - x_1)\):
\[ y - (-1) = 4(x - 4) \]
This simplifies to:
\[ y + 1 = 4x - 16 \]
Rearranging gives:
\[ 4x - y - 17 = 0 \]

### Final Equation of the Normal
Thus, the equation of the normal at the point (4, -1) is:
\[ 4x - y - 17 = 0 \]