Question 995859: Find the equation of a tangent from the point(0,5) to the circle
x^2 + y^2=16
Answer by josgarithmetic(39618)   (Show Source):
You can put this solution on YOUR website! 
The tangent from above will meet the upper half;
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
and omitting the algebraic steps here,
find that  or  .
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