Find the equation of the circle that has a diameter with endpoints located at (7,3) and (7, 5) A. (x-7)* +(4+1)2 = 16 B. (x-7)+(-1) = 4 C. (x + 1)2 +(y-7) = 16 D. (x-7)+(+1)2 = 64 Please select the best answer from the choices provided Ο Α ano a OD

Respuesta :

the equation of a circle is:

[tex](x-h)^2+(y-k)^2=r^2[/tex]

the center is (h,k) and r=radius

the center is:

[tex]\begin{gathered} x_{center}=\frac{7+7}{2}=7 \\ y_{center}=\frac{3+5}{2}=4 \end{gathered}[/tex]

h=7 k=4

and we can find the diameter as

[tex]\begin{gathered} d=\sqrt[]{(x_1-x_2)^2+(y_1-y_2)^2} \\ d=\sqrt[]{(7-7)^2+(5-3)^2} \\ d=\sqrt[]{2^2}=2 \end{gathered}[/tex]

and r=d/2

so r=1

so the equation of the circle is:

[tex](x-7)^2+(y-4)^2=1^2[/tex]