Final Answer:
The locus of the point of intersection of tangents drawn to the circles x² + y² = a² and x² + y² = b², which are perpendicular to each other, is a circle with its center at the origin and a radius of √(a² * b²).
Explanation:
To prove this, let's consider the equations of the circles x² + y² = a² and x² + y² = b². For any point (x₀, y₀) on these circles, the equation of the tangent at that point can be written as xx₀ + yy₀ = a² and xx₀ + yy₀ = b² respectively, where (x, y) is the point of tangency.
Given that the tangents drawn to these circles are perpendicular, the product of the slopes of these tangents is -1. Hence, the product of the slopes of the lines xx₀ + yy₀ = a² and xx₀ + yy₀ = b² is -1.
Considering the equations of the tangents, the product of their slopes is (1/x₀² + 1/y₀²) = -1. Solving this yields x₀² * y₀² = -1. Now, for a point (x₀, y₀) to lie on the locus of the intersection of these tangents, x₀² * y₀² must be a constant, say k.
From the above relation, x₀² * y₀² = k = a² * b². Therefore, the locus of the point of intersection of the perpendicular tangents is a circle centered at the origin (0,0) with a radius √(a² * b²). This circle is concentric with the given circles x² + y² = a² and x² + y² = b².