What is the equation of the circle whose diameter has endpoints (8,-2) and (-2,6)?

Answer:[tex](x-3)^{2} +(y-2)^{2}=41[/tex]Step-by-step explanation:step 1Find the diameter of the circlethe formula to calculate the distance between two points is equal to[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]we have[tex]A(8,-2)\\E(-2,6)[/tex]  substitute the values[tex]d=\sqrt{(6+2)^{2}+(-2-8)^{2}}[/tex][tex]d=\sqrt{(8)^{2}+(-10)^{2}}[/tex][tex]d=\sqrt{164}\ units[/tex][tex]d=2\sqrt{41}\ units[/tex]step 2Find the center of the circleThe center is the midpoint of the diameterThe center is equal to[tex]C=(\frac{8-2}{2},\frac{-2+6}{2})[/tex][tex]C=(3,2)[/tex]step 3Find the equation of the circleThe equation of the circle in center radius form is equal to[tex](x-h)^{2} +(y-k)^{2}=r^{2}[/tex]we have(h,k)=(3,2)[tex]r=2\sqrt{41}/2=\sqrt{41}\ units[/tex] ---> the radius is half the diametersubstitute[tex](x-3)^{2} +(y-2)^{2}=(\sqrt{41})^{2}[/tex][tex](x-3)^{2} +(y-2)^{2}=41[/tex]