Let's work a more general version of the problem...that is, to find the equation of the line tangent to the circle:
$$(x-h)^2+(y-k)^2=r^2$$
at the point on the circle:
$$\left(x_1,y_1\right)$$
where:
$$y_1\ne k$$
When $y_1=k$, then we know the tangent line is simply $x=x_1$.
The slope of the line will be:
$$m=\frac{h-x_1}{y_1-k}$$
And so, using the point-slope formula, the tangent line is:
$$y=\frac{h-x_1}{y_1-k}\left(x-x_1\right)+y_1$$
Arranged in slope-intercept form, we have:
$$y=\frac{h-x_1}{y_1-k}x+\frac{x_1-h}{y_1-k}x_1+y_1$$
For the given problem, we have:
$$(h,k)=(0,0)$$
And so, our formula reduces to:
$$y=-\frac{x_1}{y_1}x+\frac{x_1}{y_1}x_1+y_1=-\frac{x_1}{y_1}x+\frac{x_1^2+y_1^2}{y_1}=-\frac{x_1}{y_1}x+\frac{r^2}{y_1}$$
The first point is:
$$\left(x_1,y_1\right)=(5,12)$$
And so the tangent line is:
$$y=-\frac{5}{12}x+\frac{169}{12}$$
The second point is:
$$\left(x_1,y_1\right)=(5,-12)$$
And so the tangent line is:
$$y=\frac{5}{12}x-\frac{169}{12}$$
Let's plot the circle and the two lines to verify:
