Find an equation of the circle that has center , 1−1 and passes through , 32

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

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt

look at the picture i have filled out a table for it. hope this .