Find the locus of the point of intersections of perpendicular tangents to the circle `x^(2) +y^(2) =a^(2)`

Answer

To find the locus of the point of intersection of perpendicular tangents to the circle given by the equation \( x^2 + y^2 = a^2 \), we will follow these steps:

### Step 1: Understand the Circle Equation
The equation of the circle is given as:
\[ x^2 + y^2 = a^2 \]
This represents a circle centered at the origin (0, 0) with radius \( a \).

### Step 2: Equation of Tangent to the Circle
The equation of the tangent to the circle at a point \( P(h, k) \) is given by:
\[ x h + y k = a^2 \]
This is derived from the general equation of the tangent to a circle.

### Step 3: Perpendicular Tangents
For two tangents to be perpendicular, the product of their slopes must be -1. If we consider two tangents at points \( P_1(h_1, k_1) \) and \( P_2(h_2, k_2) \), we can express the tangents as:
1. \( x h_1 + y k_1 = a^2 \)
2. \( x h_2 + y k_2 = a^2 \)

### Step 4: Condition for Perpendicularity
The condition for the tangents to be perpendicular is:
\[ h_1 h_2 + k_1 k_2 = 0 \]
This means that the dot product of the direction vectors of the tangents is zero.

### Step 5: Locus of Intersection Point
Let \( P(h, k) \) be the point of intersection of the two tangents. The equations of the tangents can be rewritten as:
\[ x h + y k = a^2 \]
Substituting for the two tangents gives us:
\[ (x h_1 + y k_1)(x h_2 + y k_2) = (a^2)^2 \]

### Step 6: Substitute and Simplify
Using the perpendicularity condition \( h_1 h_2 + k_1 k_2 = 0 \), we can express \( k_2 \) in terms of \( h_1 \) and \( k_1 \):
\[ k_2 = -\frac{h_1 h_2}{k_1} \]
Substituting this back into the tangent equations and simplifying will lead us to a relationship involving \( h \) and \( k \).

### Step 7: Final Equation
After simplification, we arrive at the equation:
\[ h^2 + k^2 = 2a^2 \]
This represents a circle centered at the origin with radius \( \sqrt{2}a \).

### Conclusion
Thus, the locus of the point of intersection of the perpendicular tangents to the circle \( x^2 + y^2 = a^2 \) is given by:
\[ x^2 + y^2 = 2a^2 \]