The slope of the tangent at the point (h, h) of the circle `x^(2)+y^(2)=a^(2)`, is

Options

A. 0
B. 1
C. `-1`
D. Depends on h

Answer

To find the slope of the tangent at the point (h, h) of the circle defined by the equation \(x^2 + y^2 = a^2\), we can follow these steps:

### Step 1: Differentiate the equation of the circle
We start with the equation of the circle:
\[ x^2 + y^2 = a^2 \]
To find the slope of the tangent, we need to differentiate both sides with respect to \(x\).

### Step 2: Apply implicit differentiation
Differentiating both sides gives:
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(a^2) \]
This simplifies to:
\[ 2x + 2y \frac{dy}{dx} = 0 \]

### 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 (h, h)
Now, we substitute the point (h, h) into the derivative:
\[ \frac{dy}{dx} = -\frac{h}{h} \]
This simplifies to:
\[ \frac{dy}{dx} = -1 \]

### Conclusion
Thus, the slope of the tangent at the point (h, h) on the circle is:
\[ \text{slope} = -1 \]

### Final Answer
The correct option is: **C. -1**

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