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

What must I remember about Condition for two circles to intersect orthogonally?

• Orthogonal Intersection Definition: Two circles intersect orthogonally if their tangents at either point of intersection are perpendicular (angle of intersection θ = 90°).
• Equivalent Geometric Criterion: Tangent to one circle at intersection point passes through the center of the other circle, so triangle C1 P C2 is right-angled at P, giving (C1C2)² = r1² + r2².
• Condition in General Form: Let S1 ≡ x² + y² + 2g1*x + 2f1*y + c1 = 0 and S2 ≡ x² + y² + 2g2*x + 2f2*y + c2 = 0; they intersect orthogonally if and only if: 2(g1*g2 + f1*f2) = c1 + c2.
• Circle orthogonal to given circle with center on a line: Use 2(g1*g2 + f1*f2) = c1 + c2 with the linear constraint that center (-g2, -f2) satisfies line equation ax + by + c = 0.
• Radical axis and orthogonal circle: If a circle cuts two circles orthogonally, its center must lie on the radical axis of the two circles.

Test yourself (1 minute)

1. Standard derivation of 2(g1 g2 + f1 f2) = c1 + c2

   Distance squared between centers C1(-g1, -f1) and C2(-g2, -f2) is (g1 - g2)² + (f1 - f2)²; equating to r1² + r2² = (g1² + f1² - c1) + (g2² + f2² - c2) expands and simplifies directly to 2(g1 g2 + f1 f2) = c1 + c2.

2. Circle orthogonal to three given circles

   Its center is the Radical Center of the three circles, and its radius is the length of tangent from the radical center to any of the three circles.

3. Self-orthogonal circle impossibility

   A real circle cannot be orthogonal to itself since 2(g² + f²) = 2c => g² + f² - c = 0 (point circle, radius 0).

Practise Condition for two circles to intersect orthogonally questions today

Source: NIE G.C.E. (A/L) Combined Mathematics Resource Book - Pure Mathematics, Unit 19

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Year 1 · Term 3