Write the standard form of the equation of the circle with endpoints of a diameter at the points (7,2) and (-9,5).

Answer: [tex](x+1)^2 + (y- 3.5)^2 =64[/tex]Step-by-step explanation:We are given that  endpoints of a diameter at the points (7,2) and (-9,5).Center of the circle is the mid point of diameter So, first find the mid point of diameter Formula : [tex]x=\frac{x_1+x_2}{2} , y=\frac{y_1+y_2}{2}[/tex][tex](x_1,y_1)=(7,2)\\(y_1,y_2)=(-9,5)[/tex]Substitute the values in formula [tex]x=\frac{7-9}{2} , y=\frac{2+5}{2}[/tex][tex]x=-1 , y=3.5[/tex]So, center of circle = (h,k )=(-1,3.5)To find length of diameter :[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex][tex](x_1,y_1)=(7,2)\\(y_1,y_2)=(-9,5)[/tex][tex]d=\sqrt{(-9-7)^2+(5-2)^2}[/tex][tex]d=\sqrt{265}[/tex]Length of radius = r = [tex]\frac{Diameter}{2}=\frac{\sqrt{256}}{2}[/tex]Standard form of the equation of the circle : [tex](x -h)^2 + (y- k)^2 = r^2[/tex](h,k )=(-1,3.5)[tex]r=\frac{\sqrt{256}}{2}[/tex]Equation of the circle : [tex](x+1)^2 + (y- 3.5)^2 =(\frac{\sqrt{256}}{2})^2[/tex]Equation of the circle : [tex](x+1)^2 + (y- 3.5)^2 =64[/tex]Hence  the standard form of the equation of the circle with endpoints of a diameter at the points (7,2) and (-9,5) is  [tex](x+1)^2 + (y- 3.5)^2 =64[/tex]