Question

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

Answer 1

the equation of a circle is:

`(x-h)^2+(y-k)^2=r^2`

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

the center is:

`\begin{gathered} x_(center)=(7+7)/(2)=7 \\ y_(center)=(3+5)/(2)=4 \end{gathered}`

h=7 k=4

and we can find the diameter as

`\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}`

and r=d/2

so r=1

so the equation of the circle is:

`(x-7)^2+(y-4)^2=1^2`