It is given that the endpoints of the diameter of the circle are (-7,4) and (9,7)
The diameter form is given by:
[tex](x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0[/tex]Therefore it follows:
[tex]\begin{gathered} (x-(-7))(x-9)+(y-4)(y-7)=0 \\ (x+7)(x-9)+(y-4)(y-7)=0 \\ x^2+7x-9x-63+y^2-4y-7y+28=0 \\ x^2+y^2-2x-11y-35=0 \end{gathered}[/tex]Hence the equation of the circle is given by:
[tex]x^2+y^2-2x-11y-35=0[/tex]