What is the equation of this circle in standard form?

Answer:
Standard equation of circle:
[tex](x-\frac{11}{2})^2+(y-4)^2=12.25[/tex]
Step-by-step explanation:
Given: Ends points of diameter are M(2,4) and N(9,4)
Mid point of M and N is center of circle.
Let center of circle be O(h,k)
Using mid point formula:
[tex[(x,y)\rightarrow (\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})[/tex]
[tex]O(h,k)\rightarrow (\dfrac{2+9}{2},\dfrac{4+4}{2})[/tex]
[tex]O(h,k)\rightarrow (\dfrac{11}{2},4)[/tex]
Distance between O and M is radius of circle.
Let radius of circle be r
Using distance formula:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2[/tex]
[tex]r=\sqrt{(\dfrac{11}{2}-2)^2+(4-4)^2=3.5[/tex]
Standard equation of circle:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Equation of required circle:
[tex](x-\frac{11}{2})^2+(y-4)^2=3.5^2[/tex]
[tex](x-\frac{11}{2})^2+(y-4)^2=12.25[/tex]