x2 + y2 + D1x + E1y + F1 = 0 and x2 + y2 + D2x + E2y + F2 = 0
The condition for them to be orthogonal is:
D1D2 + E1E2 = 2(F1 + F2)
Now, for 2 conics:
A1x2 + B1y2 + C1xy + D1x + E1y + F1 = 0 and A1x2 + B1y2 + C1xy + D1x + E1y + F1 = 0
Deduce the condition for them to be orthogonal.
Note: Orthogonality means the curves' tangents at their points of intersection are perpendicular to each other.
更新1:
從 circles 的 orthogonality 中, 其定義為: Two circles, C1 and C2, they have 2 points of intersection P1 and P2. At P1, C1's tangent is L11, C2's tangent is L21 At P2, C1's tangent is L12, C2's tangent is L22 Result: L11 and L21 are perp., L12 and L22 are perp.
更新2:
若以只得一交點是 orthogonal 即可, 是否較易算出?
