Question

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

Answer 1

Answer:

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 = $(\frac{-2+8}{2} ,\frac{1+9}{2} )$

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

Step(ii):-

The distance of two end points

PQ = $\sqrt{(x_{2}-x_{1} )^{2} +(y_{2} -y_{1} )^{2} }$

$PQ= \sqrt{(8+2 )^{2} +(8 )^{2} }$

PQ = √164 = 12.8

Diameter    d = 2r

                 radius r = d/2

                Radius r = 6.4

Final answer:-

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