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

How do I practise Condition for two circles to intersect orthogonally?

1 question with a worked solution

Check before you answer: Condition Derivation: Expanding (g₁ - g₂)² + (f₁ - f₂)² = (g₁² + f₁² - c₁) + (g₂² + f₂² - c₂) to demonstrate 2g₁g₂ + 2f₁f₂ = c₁ + c₂.

1. What is the necessary and sufficient algebraic condition for the two circles x² + y² + 2g₁x + 2f₁y + c₁ = 0 and x² + y² + 2g₂x + 2f₂y + c₂ = 0 to intersect orthogonally (at right angles)?
   1. A) 2g₁g₂ + 2f₁f₂ = c₁ + c₂
   2. B) g₁g₂ + f₁f₂ = c₁ + c₂
   3. C) g₁/g₂ = f₁/f₂ = c₁/c₂
   4. D) 2g₁g₂ - 2f₁f₂ = c₁ - c₂
   Check my answer

   Correct answer: A) 2g₁g₂ + 2f₁f₂ = c₁ + c₂

   Two circles intersect orthogonally if the tangents at their intersection point are perpendicular, which means the radii to the intersection point form a right triangle with the line of centers: d² = r₁² + r₂². Substituting d² = (g₁ - g₂)² + (f₁ - f₂)², r₁² = g₁² + f₁² - c₁, and r₂² = g₂² + f₂² - c₂ simplifies directly to 2g₁g₂ + 2f₁f₂ = c₁ + c₂.

   Show worked solution

   Step 1: Orthogonality condition: d² = r₁² + r₂². Step 2: (g₁ - g₂)² + (f₁ - f₂)² = (g₁² + f₁² - c₁) + (g₂² + f₂² - c₂). Step 3: Expand and cancel terms: -2g₁g₂ - 2f₁f₂ = -c₁ - c₂ => 2g₁g₂ + 2f₁f₂ = c₁ + c₂. Answer: 2g₁g₂ + 2f₁f₂ = c₁ + c₂.
   Modelled on: the A/L Combined Mathematics examination pattern

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Where does Condition for two circles to intersect orthogonally fit in the syllabus?

Year 1 · Term 3