Question

Answer the following :

Tangents to the circle x² + y² = a² with inclinations, θ₁ and θ₂ intersect in P. Find the locus of such that tan θ₁ + tan θ₂ = 0

Answer 1

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

The equations of tangents with slope m to the circle x² + y² = a² are

y = `"m"x ± "a"sqrt(1 + "m"^2)` 

If these tangents pass through the point P(x₁, y₁),

we have,

y₁ = `"m"x_1 ± "a"sqrt(1 + "m"^2)` 

∴ y₁ – mx₁ = `±"a"sqrt(1 + "m"^2)`

∴ (y₁ – mx₁)² = a²(1 + m²)

∴ `"y"_1^2` – 2mx₁y₁ + `"m"^2"x"_1^2` = a² + a²m²

∴ (`"x"_1^2` – a²)m² – 2mx₁y₁ +(`"y"_1^2`  – a²) = 0

The roots m₁, m₂ of this quadratic equation in m, are the slopes of the tangents from the point P(x₁, y₁).

Also from this quadratic equation,

m₁ + m₂ = `(2x_1"y"_1)/(x_1^2 - "a"^2)`

and m₁m₂ = `(y_1^2 - "a"^2)/(x_1^2 - "a"^2)`

Since θ₁ and θ₂ are the inclinations of the tangents, tanθ₁ = m₁ and tanθ₂ = m₂.

tanθ₁ + tanθ₂ = 0

∴ m₁ + m₂ = 0

∴ `(2x_1y_1)/(x_1^2 - "a"^2)` = 0

∴ x₁y₁ = 0

∴ the equation of the locus of P(x₁, y₁) is xy = 0.