hello :<span> <span>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 </span></span>A (1 , -1) , passing through P(3, 2) is : (x-1)² + (y+1)² = 13²
