Given that end points of the diameter are A(0, 0) and B(8, 6).
Center is always at the middle of the diameter so we can use mid point formula to find the coordinate of center.
[tex] \left ( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right )
=\left ( \frac{0+8}{2}, \frac{0+6}{2} \right )
=\left ( \frac{8}{2}, \frac{6}{2} \right )
=(4,3)
[/tex]
We know that circle is given by formula
[tex] (x-h)^2+(y-k)^2=r^2 [/tex] having center at (h,k)
we found center at (4,3) so that means h=4, k=3
plug both into above formula
[tex] (x-4)^2+(y-3)^2=r^2 [/tex] ...(i)
plug the given point (0,0) into above equation
[tex] (0-4)^2+(0-3)^2=r^2 [/tex]
[tex] (-4)^2+(-3)^2=r^2 [/tex]
[tex] 16+9=r^2 [/tex]
[tex] 25=r^2 [/tex]
plug value of r^2=25 into equation (i)
[tex] (x-4)^2+(y-3)^2=25 [/tex]
Hence choice C. [tex] (x-4)^2+(y-3)^2=25 [/tex] is the final answer .
