Question 993811
Center of circle on the Origin,  {{{x^2+y^2=r^2}}}, and for this, radius is r.
You have {{{r^2=169=13^2}}}, so the radius {{{r=13}}}.


The point  ON THE CIRCLE, (5,12), is part of a tangent line which passes through this point.  This means, you want to find an equation for this line and this line TOUCHES the circle at this point; and it is perpendicular to the line which contains  this point (5,12) and the Origin (which is center of your circle.)
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Work to understand that discussion before continuing.


What is the line containing the circle's center (0,0) and the given point (5,12)?  You should find just intuitively this is  {{{y=(12/5)x}}}.   The y-intercept and the x-intercept both 0.


What is the equation for the line PERPENDICULAR to {{{y=(12/5)x}}}  and contains the point  (5,12)?  For perpendicularity, its slope must be negative reciprocal of {{{12/5}}}, so this slope needed will be {{{-5/12}}}.  You can use the point-slope equation form  (for convenience if you are comfortable with it), and plug-in the needed slope and included point (5,12):
{{{highlight(y-12=-(5/12)(x-5))}}}


Use simple algebra if you want this equation in standard form or in slope-intercept form.


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Try to make a sketch or a graph on your own to help analyze the problem description.