Question

Prosta l przecina okrąg x2+y2=20 w punktach A i B. Oblicz długość cięciwy AB: a)l:y=2x b)l:y=x+2 c)l:y=-x-6

Answer 1

a)

x^2 + y^2 = 20

y = 2x

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x^2 + (2x)^2 = 20

x^2 + 4 x^2 = 20

5 x^2 = 20 / : 5

x^2 = 4

x = - 2  lub x = 2

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zatem

y = 2*(-2) = - 4   lub   y = 2*2 = 4

czyli

A = (-2; -4),  B = ( 2; 4)

I AB I^2 = ( 2 - (-2))^2 + ( 4 - (-4))^2 = 4^2 + 8^2 = 16 + 64 = 80 = 16*5

więc

I AB I = 4 p(5)

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b)

x^2 + y^2 = 20

y =x +2

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x^2 + ( x + 2)^2 = 20

x^2 + x^2 + 4 x + 4 = 20

2 x^2 + 4 x - 16 = 0  / : 2

x^2 + 2x - 8 = 0

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delta = 2^2 - 4*1*(-8) = 4 + 32 = 36

p(delty ) = 6

x = [ - 2 - 6]/2 = - 8/2 = - 4

lub

x = [ -2 + 6 ]/2 = 4/2 =  2

więc

y = -4 + 2 = - 2   lub   y = 2 + 2 = 4

Mamy:  A = ( -4; -2)  ,  B = ( 2; 4)

I AB I^2 = ( 2 - (-4))^2 + ( 4 - (-2))^2 = 6^2 + 6^2 = 36 + 36 = 36*2

więc

I AB I = 6 p(2)

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c)

x^2 + y^2 = 20

y = - x - 6

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x^2 + ( - x - 6)^2 = 20

x^2 + x^2 + 12 x + 36 = 20

2 x^2 + 12 x + 16 = 0 / : 2

x^2 + 6 x + 8 = 0

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delta = 6^2 - 4*1*8 = 36 - 32 = 4

p(delty) = 2

x = [ -6 - 2]/2 = -8/2 = - 4

lub

x = [ - 6 + 2]/2 = -4/2 = - 2

zatem

y =  -( -4) - 6 = 4 - 6 = - 2

lub

y = - ( -2) - 6 = 2 - 6 = - 4

A = ( -4; -2) ,  B = (-2; - 4)

więc

I AB I^2 = ( - 2 - (-4))^2 + ( -4 - (-2))^2 = 2^2 + (-2)^2 = 4 + 4 = 4*2

czyli

I AB I = 2 p(2)

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p(2) - pierwiastek kwadratowy z  2

p(5) - pierwiastek kwadratowy z  5