Solution:
Given:
[tex]\begin{gathered} center\text{ }=(0,-1) \\ Through\text{ p}oint\text{ }(-35,0) \end{gathered}[/tex]The equation of a circle is gotten by;
[tex]\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ where: \\ x=-35 \\ y=0 \\ h=0 \\ k=-1 \\ \end{gathered}[/tex]Substituting these values into the equation to get the value of r;
[tex]\begin{gathered} (-35-0)^2+(0-(-1))^2=r^2 \\ (-35)^2+(1)^2=r^2 \\ 1225+1=r^2 \\ r^2=1226 \end{gathered}[/tex]Thus, the equation of the circle is;
[tex]\begin{gathered} (x-0)^2+(y-(-1))^2=1226 \\ x^2+(y+1)^2=1226 \end{gathered}[/tex]