Solution:
Given the center O(h,k), of a circle and a point A(a,b), it is passing through.
The distance from the center to the circumference is called the radius, r. Thus;
[tex]\begin{gathered} O(0,-1),A(\sqrt{35},0) \\ \\ |OA|=r=\sqrt{(0-(-1))^2+(\sqrt{35}-0)^2} \\ \\ r=\sqrt{1+35} \\ \\ r=6 \\ \end{gathered}[/tex]
Since the equation of a circle is;
[tex](x-h)^2+(y-k)^2=r^2[/tex]
The, the equation of the circle is;
[tex]\begin{gathered} (h,k)=(0,-1) \\ \\ (x-0)^2+(y-(-1))^2=6^2 \\ \\ x^2+(y+1)^2=36 \end{gathered}[/tex]
ANSWER:
[tex]x^{2}+(y+1)^{2}=36[/tex]
