IF the slope of the tangent to the circle `S=x^2+y^2-13=0` at (2,3) is m, then the point `(m,(-1)/m)` is

Options

A. an external point with respect to the circle S=0
B. an internal point with respect to the circle S=0
C. the centre of the circle S=0
D. a point on the circle S=0

Answer

To solve the problem step by step, we will follow the instructions given in the video transcript.

### Step 1: Identify the equation of the circle
The equation of the circle is given as:
\[ S: x^2 + y^2 - 13 = 0 \]

### Step 2: Differentiate the equation of the circle
To find the slope of the tangent line at a specific point on the circle, we need to differentiate the equation with respect to \( x \).
Differentiating both sides:
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(13) = 0 \]
This gives:
\[ 2x + 2y \frac{dy}{dx} = 0 \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ 2y \frac{dy}{dx} = -2x \]
\[ \frac{dy}{dx} = -\frac{x}{y} \]

### Step 4: Substitute the point (2, 3) into the derivative
Now, we will find the slope \( m \) at the point \( (2, 3) \):
\[ m = \frac{dy}{dx} \bigg|_{(2, 3)} = -\frac{2}{3} \]

### Step 5: Find the coordinates of the point \((m, -\frac{1}{m})\)
Now, we need to find the point \( (m, -\frac{1}{m}) \):
\[ m = -\frac{2}{3} \]
\[ -\frac{1}{m} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} \]
Thus, the point is:
\[ \left(-\frac{2}{3}, \frac{3}{2}\right) \]

### Step 6: Determine if the point is inside or outside the circle
To find out if the point \( \left(-\frac{2}{3}, \frac{3}{2}\right) \) is inside or outside the circle, we substitute these coordinates into the circle's equation:
\[ S = \left(-\frac{2}{3}\right)^2 + \left(\frac{3}{2}\right)^2 - 13 \]
Calculating each term:
\[ \left(-\frac{2}{3}\right)^2 = \frac{4}{9}, \quad \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]
Now, convert \(\frac{9}{4}\) to a fraction with a denominator of 36:
\[ \frac{9}{4} = \frac{81}{36} \]
Convert \(\frac{4}{9}\) to a fraction with a denominator of 36:
\[ \frac{4}{9} = \frac{16}{36} \]
Now, substituting back into \( S \):
\[ S = \frac{16}{36} + \frac{81}{36} - 13 = \frac{97}{36} - 13 \]
Convert 13 to a fraction with a denominator of 36:
\[ 13 = \frac{468}{36} \]
So,
\[ S = \frac{97}{36} - \frac{468}{36} = \frac{97 - 468}{36} = \frac{-371}{36} \]
Since \( S < 0 \), the point is inside the circle.

### Final Answer
The point \( \left(-\frac{2}{3}, \frac{3}{2}\right) \) is inside the circle.

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