The points on the hyperbola `x^(2)/a^(2) - y^(2)/b^(2) = 1` from where mutually perpendicular tangents can be drawn to circle `x^(2) + y^(2) = a^(2)/2` is /are

Options

A. `(a sqrt(3/2), a/sqrt2)`
B. `(-a sqrt(3/2) , a/sqrt2)`
C. `(-a sqrt(3/2) , - a/sqrt2)`
D. `(a sqrt(3/2, - a/sqrt2)) `

Answer

To find the points on the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) from where mutually perpendicular tangents can be drawn to the circle \( x^2 + y^2 = \frac{a^2}{2} \), we can follow these steps:

### Step 1: Understand the condition for mutually perpendicular tangents
The condition for mutually perpendicular tangents drawn from a point to a circle is that the point lies on the auxiliary circle of the circle. The equation of the auxiliary circle for the circle \( x^2 + y^2 = r^2 \) is given by \( x^2 + y^2 = 2r^2 \).

### Step 2: Find the auxiliary circle for the given circle
Here, the radius \( r \) of the circle \( x^2 + y^2 = \frac{a^2}{2} \) is \( r = \sqrt{\frac{a^2}{2}} = \frac{a}{\sqrt{2}} \). Therefore, the equation of the auxiliary circle is:
\[ x^2 + y^2 = 2 \left( \frac{a^2}{2} \right) = a^2 \]

### Step 3: Set up the equations
We need to find points \( (x, y) \) on the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) that also satisfy the equation of the auxiliary circle \( x^2 + y^2 = a^2 \).

### Step 4: Substitute and simplify
From the hyperbola equation, we can express \( y^2 \) in terms of \( x^2 \):
\[ y^2 = b^2 \left( \frac{x^2}{a^2} - 1 \right) \]
Substituting this into the auxiliary circle equation:
\[ x^2 + b^2 \left( \frac{x^2}{a^2} - 1 \right) = a^2 \]
This simplifies to:
\[ x^2 + \frac{b^2}{a^2} x^2 - b^2 = a^2 \]
Combining like terms:
\[ \left(1 + \frac{b^2}{a^2}\right)x^2 = a^2 + b^2 \]
\[ \frac{(a^2 + b^2)}{a^2} x^2 = a^2 + b^2 \]
Thus,
\[ x^2 = a^2 \quad \Rightarrow \quad x = a \text{ or } x = -a \]

### Step 5: Find corresponding \( y \) values
Substituting \( x = a \) into the hyperbola equation:
\[ \frac{a^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \Rightarrow \quad 1 - \frac{y^2}{b^2} = 1 \quad \Rightarrow \quad y^2 = 0 \quad \Rightarrow \quad y = 0 \]
So one point is \( (a, 0) \).

Now substituting \( x = -a \):
\[ \frac{(-a)^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \Rightarrow \quad 1 - \frac{y^2}{b^2} = 1 \quad \Rightarrow \quad y^2 = 0 \quad \Rightarrow \quad y = 0 \]
So the other point is \( (-a, 0) \).

### Final Answer
The points on the hyperbola from where mutually perpendicular tangents can be drawn to the circle are:
\[ (a, 0) \text{ and } (-a, 0) \]