Respuesta :

 

hello :
an equation of the circle Center at the A(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
in this exercice : a = 1  and b = -1 (Center at A (1 , -1) 
r = AP
r² = (AP)²    ....... P(3,2)
r² = (3-1)² +(2+1)² = 4+9=13
an equation of the circle that satisfies the stated conditions.
Center at 
A (1 , -1)  , passing through P(3, 2) is :  (x-1)² + (y+1)² = 13²