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Posição relativa entre duas circunferências

Posição relativa entre duas circunferências. Secante Tangente Distintas (sem pontos comuns). 5. 5. 5. -. λ 1 : x 2 + y 2 = 30 λ 2 : (x – 3) 2 + y 2 = 9. =>. y 2 = 30 – x 2. (x – 3) 2 + (30 – x 2 ) = 9 (x 2 – 6x + 9) + (30 – x 2 ) – 9 = 0 6x + 30 = 0 30 = 6x x = 5.

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Posição relativa entre duas circunferências

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  1. Posição relativa entre duas circunferências Secante Tangente Distintas (sem pontos comuns)

  2. 5 5 5 - • λ1: x2 + y2 = 30 • λ2: (x – 3)2 + y2 = 9 => y2 = 30 – x2 • (x – 3)2 + (30 – x2) = 9 • (x2 – 6x + 9) + (30 – x2) – 9 = 0 • 6x + 30 = 0 • 30 = 6x • x = 5 Substituindo x = 5 x2 + y2 = 30 (5)2 + y2 = 30 25 + y2 = 30 y2 = 30 – 25 y = • ± 5 Secantes •

  3. • • C1 C2 dC1,C2 r1 + r2 ≠ Substituindo x = 11 x2 + y2 – 2x – 2y – 98 = 0 (11)2 + y2 – 2(11) – 2y – 98 = 0 y2 – 2y + 1 = 0 - 2) λ1: x2 + y2 – 20x – 2y + 100 = 0 λ2: x2 + y2 – 2x – 2y – 98 = 0 • 18x + 198 = 0 • 198 = 18x • x = 198/18 • x = 11 Δ = b2 – 4ac Δ = 0 y = 1 r1 r2 1 1 • • 11 11 Tangente externa Tangente Tangente interna dC1,C2 r1 + r2 =

  4. 2) λ1: x2 + y2 = 1 λ2: (x + 2)2 + (y – 2)2 = 1 r1 • C1 • C2 r2 x2 + y2 = (x + 2)2 + (y – 2)2 x2 + y2 = (x2 + 4x + 4) + (y2 – 4y + 4) x2 + y2 - x2 - 4x - 4 - y2 + 4y - 4 = 0 - 4x + 4y - 8 = 0 Interna __ __ __ __ 4 4 4 4 -x + y - 2 = 0 y = x + 2 d c1,c2 r1 + r2 < ? r2 Distintas (nenhum ponto comum) • C2 r1 • C1 Externas • C1 = C2 d c1,c2 r1 + r2 > Concêntrica

  5. Página 572exercícios 25 até 28

  6. 25) Dadas as circunferências, descubra suas posições relativas e seus pontos comuns (se houver): 1) λ1: x2 + y2 – 4x – 8y – 5 = 0 λ2: x2 + y2 – 2x – 6y + 1 = 0 2) λ1:(x – 2)2 + (y – 1)2 = 4 λ2: (x – 2)2 + (y + 2)2 = 1

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