If `x/alpha+y/beta=1` touches the circle `x^2+y^2=a^2` then point `(1/alpha , 1/beta)` lies on (a) straight line (b) circle (c) parabola (d) ellipse

Options

A. a straight line
B. a circle
C. a parabola
D. an ellipse

Explanation

To solve the problem step by step, we need to analyze the given information and derive the required result.

### Step 1: Understand the Given Information
We have a line given by the equation:
\[ \frac{x}{\alpha} + \frac{y}{\beta} = 1 \]
This line touches the circle defined by the equation:
\[ x^2 + y^2 = a^2 \]
We need to determine the locus of the point \(\left(\frac{1}{\alpha}, \frac{1}{\beta}\right)\).

### Step 2: Write the Tangent Equation for the Circle
The point form of the tangent to the circle \(x^2 + y^2 = a^2\) at the point \((x_1, y_1)\) is given by:
\[ x_1 x + y_1 y = a^2 \]
Since the line touches the circle, we can equate the two expressions.

### Step 3: Relate the Line and Tangent Equation
From the line equation, we can express \(x_1\) and \(y_1\) in terms of \(\alpha\) and \(\beta\):
\[ \frac{x_1}{\alpha} = \frac{y_1}{\beta} = a^2 \]
From this, we can derive:
\[ x_1 = a^2 \alpha \quad \text{and} \quad y_1 = a^2 \beta \]

### Step 4: Substitute \(x_1\) and \(y_1\) into the Circle Equation
Since \((x_1, y_1)\) lies on the circle, we substitute these values into the circle's equation:
\[ (x_1)^2 + (y_1)^2 = a^2 \]
Substituting \(x_1\) and \(y_1\):
\[ (a^2 \alpha)^2 + (a^2 \beta)^2 = a^2 \]
This simplifies to:
\[ a^4 \alpha^2 + a^4 \beta^2 = a^2 \]

### Step 5: Simplify the Equation
Dividing the entire equation by \(a^2\):
\[ a^2 \alpha^2 + a^2 \beta^2 = 1 \]
This can be rewritten as:
\[ \frac{1}{a^2} = \frac{1}{\alpha^2} + \frac{1}{\beta^2} \]

### Step 6: Identify the Locus
Let \(h = \frac{1}{\alpha}\) and \(k = \frac{1}{\beta}\). Then we can rewrite the equation as:
\[ h^2 + k^2 = \frac{1}{a^2} \]
This represents the equation of a circle centered at the origin \((0, 0)\) with radius \(\frac{1}{a}\).

### Conclusion
Thus, the locus of the point \(\left(\frac{1}{\alpha}, \frac{1}{\beta}\right)\) is a circle.

### Final Answer
The correct option is (b) circle.

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