If the line hx + ky = 1 touches `x^(2)+y^(2)=a^(2)`, then the locus of the point (h, k) is a circle of radius

Options

A. a
B. 1/a
C. `sqrt(a)`
D. `1//sqrt(a)`

Explanation

To solve the problem, we need to find the locus of the point (h, k) such that the line \( hx + ky = 1 \) is tangent to the circle \( x^2 + y^2 = a^2 \).

### Step-by-Step Solution:

1. **Identify the Given Equations**:
- The equation of the line is \( hx + ky = 1 \).
- The equation of the circle is \( x^2 + y^2 = a^2 \).

2. **Determine the Condition for Tangency**:
- For the line to be tangent to the circle, the distance from the center of the circle (which is at the origin (0,0)) to the line must be equal to the radius of the circle (which is \( a \)).
- The formula for the distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
- Here, \( A = h \), \( B = k \), and \( C = -1 \), and the point is \( (0, 0) \).

3. **Calculate the Distance**:
- Substitute into the distance formula:
\[ d = \frac{|h(0) + k(0) - 1|}{\sqrt{h^2 + k^2}} = \frac{|-1|}{\sqrt{h^2 + k^2}} = \frac{1}{\sqrt{h^2 + k^2}} \]

4. **Set the Distance Equal to the Radius**:
- Since the distance must equal the radius \( a \):
\[ \frac{1}{\sqrt{h^2 + k^2}} = a \]

5. **Square Both Sides**:
- Squaring both sides gives:
\[ \frac{1}{h^2 + k^2} = a^2 \]

6. **Rearranging the Equation**:
- Rearranging this equation leads to:
\[ h^2 + k^2 = \frac{1}{a^2} \]

7. **Identify the Locus**:
- The equation \( h^2 + k^2 = \frac{1}{a^2} \) represents a circle centered at the origin with radius \( \frac{1}{a} \).

### Conclusion:
Thus, the locus of the point \( (h, k) \) is a circle with radius \( \frac{1}{a} \).