An equation of the normal at the point (2,3)to the circle `x ^(2)+ y^(2) - 2x - 2y - 3=0` is

Options

A. `2x + y - 7=0`
B. `x + 2y - 3=0`
C. `2x-y-1 =0`
D. `x - 2y + 1=0`

Answer

To find the equation of the normal at the point (2, 3) to the circle given by the equation \( x^2 + y^2 - 2x - 2y - 3 = 0 \), we can follow these steps:

### Step 1: Identify the center and radius of the circle
The general form of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the given equation, we can rewrite it as:
\[ x^2 + y^2 - 2x - 2y - 3 = 0 \]
Here, \( 2g = -2 \) and \( 2f = -2 \), which gives us:
\[ g = -1, \quad f = -1 \]
The center of the circle \( (h, k) \) is given by:
\[ (h, k) = (-g, -f) = (1, 1) \]
The radius \( r \) can be found using the formula:
\[ r = \sqrt{g^2 + f^2 - c} = \sqrt{(-1)^2 + (-1)^2 - (-3)} = \sqrt{1 + 1 + 3} = \sqrt{5} \]

### Step 2: Find the slope of the radius at the point (2, 3)
The slope of the radius from the center (1, 1) to the point (2, 3) is calculated as follows:
\[ \text{slope of radius} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 1}{2 - 1} = \frac{2}{1} = 2 \]

### Step 3: Find the slope of the normal
The normal at a point on the circle is perpendicular to the radius at that point. Therefore, the slope of the normal \( m \) is the negative reciprocal of the slope of the radius:
\[ m = -\frac{1}{\text{slope of radius}} = -\frac{1}{2} \]

### Step 4: Use the point-slope form to write the equation of the normal
Using the point-slope form of the equation of a line, which is given by:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = (2, 3) \) and \( m = -\frac{1}{2} \):
\[ y - 3 = -\frac{1}{2}(x - 2) \]

### Step 5: Simplify the equation
Now, we simplify the equation:
\[ y - 3 = -\frac{1}{2}x + 1 \]
Adding 3 to both sides:
\[ y = -\frac{1}{2}x + 4 \]
To express it in standard form, we can rearrange it:
\[ \frac{1}{2}x + y - 4 = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ x + 2y - 8 = 0 \]

### Final Answer
Thus, the equation of the normal at the point (2, 3) to the circle is:
\[ x + 2y - 8 = 0 \]