Question

Find the locus of a point, the tangents from which to the circle x² + y² = a² are mutually perpendicular

Answer 1

Let P(x1, y1) be any point on the locus.

Equation of a tangent with slope ‘m’ to the circle x² + y² = a² is y = mx ± a`sqrt(1+m^2)`

This tangent passes through P(x₁, y₁)

∴ y₁ = mx₁ ± a `sqrt(1+m^2)`

∴ (y₁ – mx₁)² = a²(1 + m²)

∴ `(x_1^2 − a^2)m^2 − 2x_1y_1m + (y_1^2 − a^2)` = 0 ….(i)

This is a quadratic equation in ‘m’.

Let m1 and m2 be slopes of two tangents drawn from P(x1, y1) to the circle.

Thus, it has two roots say m1 and m2, which are the slopes of tangents drawn from P.

∴ `m_1.m_2 = (y_1^2 - a^2)/(x_1^2  -a^2)`

∴ `y_1^2 − a^2 = −x_1^2 + a^2`

∴ `x_1^2 + y_1^2 = 2a^2`

∴ the equation of the locus of P(x₁, y₁) is x² + y² = 2a².