PARTE 2

Igualamos las distancias:

A(2;3)
r

C
r
B ( 6 ; -2 )

d (C ; A) = d (C ; B )
√[(3)2 + ( 2 – h ) 2 ] = √ [( 6 – h ) 2 + (-2) 2 ]
√[ 9 + 4 – 4h + h 2 ] = √ [ 36 – 12h + h 2 + 4 ]
(√ [ h 2 - 4h +13 ]) 2 = (√ [ h 2 - 12h + 40]) 2
h 2 - 4h +13 = h 2 - 12h + 40
8h = 27
h = 27/8

h = 27/8

k=0
PARTE 2

Hallamos el radio:
d => √[( 2 – 27/8)2 + ( 3 - 0) 2] = r
√[(- 11/64) 2 + 9] = r
√ (121/64 + 9) = r
√ 697/64= r

A(2;3)
r

C
r
B ( 6 ; -2 )

Hallamos la ecuación ordinaria:
( x – 27/4) 2 + y 2 = 697/64

h = 27/8

k=0
r = √ 697/64

Mi correccion

  • 1.
    PARTE 2 Igualamos lasdistancias: A(2;3) r C r B ( 6 ; -2 ) d (C ; A) = d (C ; B ) √[(3)2 + ( 2 – h ) 2 ] = √ [( 6 – h ) 2 + (-2) 2 ] √[ 9 + 4 – 4h + h 2 ] = √ [ 36 – 12h + h 2 + 4 ] (√ [ h 2 - 4h +13 ]) 2 = (√ [ h 2 - 12h + 40]) 2 h 2 - 4h +13 = h 2 - 12h + 40 8h = 27 h = 27/8 h = 27/8 k=0
  • 2.
    PARTE 2 Hallamos elradio: d => √[( 2 – 27/8)2 + ( 3 - 0) 2] = r √[(- 11/64) 2 + 9] = r √ (121/64 + 9) = r √ 697/64= r A(2;3) r C r B ( 6 ; -2 ) Hallamos la ecuación ordinaria: ( x – 27/4) 2 + y 2 = 697/64 h = 27/8 k=0 r = √ 697/64