1. 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

2. 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