Question 995859
{{{y^2=-x^2+16}}}
{{{y^2=16-x^2}}}
{{{y=0+- sqrt(16-x^2)}}}
The tangent from above will meet the upper half;
{{{y=sqrt(16-x^2)}}}


Refer to three different points.  Center of circle is the origin, (0,0);  the given point is (0,5) and is above the circle on the y-axis; and the general point ON the upper branch for the function of the circle,  (x,sqrt(16-x^2)).  There will be two such general points on the circle, one being for x less than 0 and another for x being greater than 0.


Make a drawing of the description...


What you want for a line containing (0,5) to be tangent with the circle, is the slopes of line containing  (0,0) and (x,sqrt(16-x^2)); and line containing  (0,5) and (x,sqrt(16-x^2)), to be negative reciprocals of each other.  Think, "radius", which will touch the circle at the tangent point(s).


Formulas for slope, both lines, their product of slopes be negative 1; and then just solve for x.


After starting the algebra,  the equation to setup is 
{{{(sqrt(16-x^2)/x)((sqrt(16-x^2)-5)/x)=-1}}}
and omitting the algebraic steps here,
find that {{{highlight(x=-12/5)}}}  or {{{highlight(x=12/5)}}}.