Answer
The equation of the circle is $x^2+y^2=169$, and the line's equation is $y={5\over12}x+{169\over12}$
Work Step by Step
The standard equation of a circle is $(x-h)^2+(y-k)^2=r^2$ where $(h, k)$ is the center of the circle and $r$ is the radius. Since the center is at the origin, $h$ and $k$ are both $0$. To find the radius, we'll use the equation $d=\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2}$ to find the distance between two points, in this case, $(0, 0)$ and $(-5, 12)$.
$d=\sqrt{(-5-0)^2+(12-0)^2} = \sqrt{25+144}$
$d=\sqrt{169} = 13$
Now we can fill in the standard equation with the values found:
$(x-0)^2+(y-0)^2=13^2$ --> $x^2+y^2=169$
To find the line's equation, we have to find its slope. To do so, we'll find the slope of its perpendicular line, which is the line that passes through the points $(0,0)$ and $(-5,12)$. We can see that it has a slope $m_{p}=-{12\over 5}$.
To calculate the slope of a perpendicular line, we take its negative inverse: $m=-{1\over m_{p}}=-{1\over -{12\over5}} = {5\over 12}$
And that will be the slope of the line that is tangent to the circle. Now, we'll use the point $(-5,12)$ and the slope $m$ in the standard form $y=mx+b$ to find $b$ and then the line's equation.
$12={5\over 12} \cdot (-5)+b$ --> ${144\over 12}+{25\over 12}=b={169\over12}$
$y={5\over12}x+{169\over12}$
