Question 895166
Since it passes through the origin (0,0) is a point on the circle.
{{{(x-0)^2+(y-3)^2=R^2}}}
Substituting,
{{{(0-0)^2+(0-3)^2=R^2}}}
{{{R^2=9}}}
So then,
{{{highlight(x^2+(y-3)^2=9)}}}
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{{{drawing(300,300,-6,6,-2,10,grid(1),blue(circle(0,3,3)),circle(0,3,0.1))}}}