The equation of the normal to the circle `x^2+y^2=9` at the point `(1/sqrt2 ,1/sqrt2)` is

Options

A. `x-y=sqrt2/3`
B. x+y=0
C. x-y=0
D. none of these

Answer

To find the equation of the normal to the circle \( x^2 + y^2 = 9 \) at the point \( \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \), we can follow these steps:

### Step 1: Find the derivative of the circle's equation
The equation of the circle is given by:
\[ x^2 + y^2 = 9 \]
To find the slope of the tangent line at the point, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(9) \]
This gives us:
\[ 2x + 2y \frac{dy}{dx} = 0 \]
Rearranging this, we find:
\[ 2y \frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y} \]

### Step 2: Evaluate the derivative at the given point
We need to find the slope of the tangent line at the point \( \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \):
\[ \frac{dy}{dx} \bigg|_{\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)} = -\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = -1 \]
Thus, the slope of the tangent line at this point is \( -1 \).

### Step 3: Find the slope of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore:
\[ \text{slope of normal} = -\frac{1}{-1} = 1 \]

### Step 4: Use the point-slope form to find the equation of the normal
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \) and \( m = 1 \):
\[ y - \frac{1}{\sqrt{2}} = 1 \left( x - \frac{1}{\sqrt{2}} \right) \]
Simplifying this, we get:
\[ y - \frac{1}{\sqrt{2}} = x - \frac{1}{\sqrt{2}} \]
\[ y = x \]

### Step 5: Write the final equation
The equation of the normal line can also be expressed in the standard form:
\[ x - y = 0 \]

Thus, the equation of the normal to the circle at the given point is:
\[ \boxed{x - y = 0} \]