Question 1038686
So the general equation of a circle is,
{{{(x-h)^2+(y-k)^2=R^2}}}
So we know,
{{{R=13}}}
{{{k=-12}}}
Substituting,
{{{(x-h)^2+(y+12)^2=169}}}
Using the point,
{{{(0-h)^2+(0+12)^2=169}}}
{{{h^2+144=169}}}
{{{h^2=25}}}
{{{h=0 +- 5}}}
There are two solutions,
{{{(x-5)^2+(y-12)^2=169}}}
and
{{{(x+5)^2+(y-12)^2=169}}}
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*[illustration circl1.JPG].