Show that the tangents of the circle (x − 6) ^2 + (y + 2)^2 = 20 passing through the origin are perpendicular to each other

Question

Show that the tangents of the circle (x − 6) ^2 + (y + 2)^2 = 20 passing through the origin are perpendicular to each other

Answer 1

y = mx; (x - 6)² + (y + 2)² = 20; 2(x - 6) + 2(y + 2)y' = 0

x - 6 + (mx + 2)m = 0;

x + m²x = 6 - 2m; x = (6 - 2m)/(1 + m²); ((6 - 2m)/(1 + m²) - 6)² + (m(6 - 2m)/(1 + m²) + 2)² = 20;

(-2m - 6m²)² + (6m - 2m² + 2 + 2m²)² = 20(1 + m²)²;

4m² + 24m³ + 36m⁴ + 36m² + 24m + 4 = 20 + 40m² + 20m⁴; 16m⁴ + 24m³ + 24m - 16 = 0;

2m⁴ + 3m³ + 3m - 2 = 0; 2(m + 1)(m - 1)(m² + 1) + 3m(m² + 1) = 0. Reducing by m² + 1 ≠ 0;

2m² + 3m - 2 = 0; (2m - 1)(m + 2) = 0; m₁ = - 2 and m₂ = 1/2

So, m₁·m₂ = -1 so this tangent lines arev perpendicular