Find an equation of the circle that satisfies the given conditions. Endpoints of a diameter are P(−2, 1) and Q(8, 9)

Equation of the circle   (x-3)²+(y-5)²=(6.4)²

                             x² -6x +9 +y² -10y +25 = 40.96

Step-by-step explanation:

Step(i):-

Given endpoints of diameter P(−2, 1) and Q(8, 9)

Centre of circle = midpoint of diameter

                   Centre =

               Centre (h, k) = (3 , 5)

Step(ii):-

The distance of two end points

PQ =

PQ = √164 = 12.8

Diameter    d = 2r

                 radius r = d/2

                Radius r = 6.4

Final -

Equation of the circle  

                    (x-h)²+(y-k)² = r²

                   (x-3)²+(y-5)²=(6.4)²

x² -6x +9 +y² -10y +25 = 40.96

x² -6x  +y² -10y  = 40.96-34

x² -6x  +y² -10y -7= 0

there are no short passages to look at

answer: need picture of problem

step-by-step explanation: