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ROVNICE TEČNY KE KRUŽNICI. rovnice tečny ke kružnici (x – x S ) 2 + (y – y S ) 2 = r 2 v tečném bodě T = [x T , y T ] se ur čí ze vztahu:. (x – x S ) 2 + (y – y S ) 2 = r 2. (x – x S ) . (x – x S ) + (y – y S ). (y – y S ) = r 2.

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slide2

ROVNICE TEČNY KE KRUŽNICI

  • rovnice tečny ke kružnici (x – xS)2 + (y – yS)2 = r2 vtečném
  • bodě T = [xT, yT] se určí ze vztahu:

(x – xS)2 + (y – yS)2 = r2

(x – xS) . (x – xS) + (y – yS). (y – yS) = r2

(xT – xS) . (x – xS) + (yT – yS). (y – yS) = r2

  • tento vztah se pak upraví na obecnou rovnici přímky
slide3

ROVNICE TEČNY KE KRUŽNICI

Příklad 1: Určete rovnici tečny ke kružnici

(x – 5)2 + (y + 3)2 = 26 v bodě T = [6, yT> 0]

- musíme určit y-ovou souřadnici tečného bodu

T є k: (6 – 5)2 + (yT + 3)2 = 26

1 + yT2 +6yT + 9 = 26

yT2 +6yT – 16 = 0

D = 62 – 4.(-16) = 100

T = [6, 2]

y1 = 2, y2 = -8

slide4

ROVNICE TEČNY KE KRUŽNICI

Příklad 1: Určete rovnici tečny ke kružnici

(x – 5)2 + (y + 3)2 = 26 v bodě T = [6, 2]

(x – 5) . (x – 5) + (y +3). (y +3) = 26

(xT – 5) . (x – 5) + (yT+3). (y +3) = 26

(6 – 5) . (x – 5) + (2+3). (y +3) = 26

x – 5 + 5y +15 = 26

t: x + 5y – 16 = 0

slide5

ROVNICE TEČNY KE KRUŽNICI

Příklad 2: Určete rovnici tečny ke kružnici

x2 + y2 + 4x – 8y – 20 = 0 v bodě T = [4, 6]

x2 + 4x + 4 – 4 + y2 – 8y + 16 – 16 – 20 = 0

(x + 2)2 – 4 + (y – 4)2 – 16 – 20 = 0

k: (x + 2)2 + (y – 4)2 = 40

(xT + 2). (x + 2) + (yT – 4). (y – 4) = 40

(x + 2). (x + 2) + (y – 4). (y – 4) = 40

(4 + 2). (x + 2) + (6 – 4). (y – 4) = 40

6x + 12 + 2y – 8 = 40

6x + 2y – 36 = 0

t: 3x + y – 18 = 0

slide6

ROVNICE TEČNY KE KRUŽNICI

Příklad 3: Určete rovnici tečny ke kružnici

x2 + y2 – 2x + 6y – 22 = 0 v bodě T = [xT< 0, 1]

x2– 2x + 1 – 1 + y2+6y + 9 – 9 – 22 = 0

(x – 1)2 – 1 + (y +3)2 – 9 – 22 = 0

k: (x – 1)2 + (y + 3)2 = 32

x2+ 1 – 2x + 6 – 22 = 0

x2 – 2x – 15 = 0

D = (-2)2 – 4.(-15) = 64

x1 = 5, x2 = -3

T = [-3, 1]

slide7

ROVNICE TEČNY KE KRUŽNICI

Příklad 3: Určete rovnici tečny ke kružnici

x2 + y2 – 2x + 6y – 22 = 0 v bodě T = [xT< 0, 1]

k: (x – 1)2 + (y + 3)2 = 32

T = [-3, 1]

(xT– 1). (x – 1) + (yT+ 3). (y + 3) = 32

(x – 1). (x – 1) + (y + 3). (y + 3) = 32

(-3 – 1). (x – 1) + (1 + 3). (y + 3) = 32

-4x + 4 + 4y + 12 = 32

-4x + 4y – 16 = 0

t: x – y + 4 = 0