Condition for two circles to intersect orthogonally — A/L Combined Mathematics

Two circles are said to intersect orthogonally if the tangents at either point of intersection are mutually perpendicular. Since the tangent to a circle is perpendicular to its radius at the point of contact, this implies that the tangent to one circle passes through the centre of the other circle. Applying Pythagoras' theorem to the right-angled triangle formed by the two centres and an intersection point gives: C₁C₂² = r₁² + r₂². Expressed in terms of the general equations S₁ ≡ x² + y² + 2g₁x + 2f₁y + c₁ = 0 and S₂ ≡ x² + y² + 2g₂x + 2f₂y + c₂ = 0, this geometric condition simplifies to the fundamental algebraic criterion: 2g₁g₂ + 2f₁f₂ = c₁ + c₂.

Check before you answer (A/L Pure Maths Paper 1)

1. Condition Derivation: Expanding (g₁ - g₂)² + (f₁ - f₂)² = (g₁² + f₁² - c₁) + (g₂² + f₂² - c₂) to demonstrate 2g₁g₂ + 2f₁f₂ = c₁ + c₂.
2. Orthogonality with Unknown Parameters: Using the condition to set up linear equations in unknown coefficients (g, f, c) of a variable circle.
3. Geometrical Implication: Proving that if a circle cuts two fixed circles orthogonally, its centre lies on their common radical axis.