Number of points on hyperbola `(x ^(2))/(a ^(2)) - (y ^(2))/(b ^(2)) =1` from were mutually perpendicular tangents can be drawn to circel `x ^(2) + y ^(2)=a ^(2)` is

Options

A. 2
B. 3
C. infinite
D. 4

Answer

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

### Step 1: Identify the auxiliary circle
The auxiliary circle of the given circle \(x^2 + y^2 = a^2\) is given by the equation \(x^2 + y^2 = 2a^2\). This circle represents the locus of points from which tangents can be drawn to the original circle.

**Hint:** The auxiliary circle is derived from the original circle by scaling the radius.

### Step 2: Set up the equations
We have two equations:
1. The hyperbola: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
2. The auxiliary circle: \(x^2 + y^2 = 2a^2\)

**Hint:** We need to find the points of intersection between these two curves.

### Step 3: Substitute and simplify
From the equation of the auxiliary circle, we can express \(y^2\) in terms of \(x^2\):
\[ y^2 = 2a^2 - x^2 \]
Now, substitute \(y^2\) into the hyperbola equation:
\[ \frac{x^2}{a^2} - \frac{2a^2 - x^2}{b^2} = 1 \]

**Hint:** This substitution allows us to eliminate \(y\) and work with a single variable.

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ \frac{x^2}{a^2} + \frac{x^2}{b^2} = 1 + \frac{2a^2}{b^2} \]
Multiply through by \(a^2b^2\) to eliminate the denominators:
\[ b^2x^2 + a^2x^2 = a^2b^2 + 2a^4 \]
This simplifies to:
\[ (b^2 + a^2)x^2 = a^2b^2 + 2a^4 \]

**Hint:** This step leads to a quadratic equation in \(x^2\).

### Step 5: Solve for \(x^2\)
Now, we can solve for \(x^2\):
\[ x^2 = \frac{a^2b^2 + 2a^4}{a^2 + b^2} \]
This gives us the values of \(x^2\) at the intersection points.

**Hint:** The number of solutions for \(x^2\) will determine the number of intersection points.

### Step 6: Determine the number of points
Since the hyperbola is symmetric and we can have positive and negative values for \(x\), each valid \(x^2\) will correspond to two \(x\) values (one positive and one negative). Additionally, for each \(x\), there will be two corresponding \(y\) values (positive and negative) from the auxiliary circle equation.

Thus, if we find \(n\) valid \(x^2\) values, the total number of points is \(4n\).

**Hint:** Count the number of valid \(x^2\) solutions to find the total number of intersection points.

### Conclusion
After solving, we find that there are 4 intersection points between the hyperbola and the auxiliary circle. Therefore, the number of points on the hyperbola from which mutually perpendicular tangents can be drawn to the circle is:

**Final Answer:** 4