Question 1018000
(x+5)2+(y-2)2=25
change to 
(x+5)^2+(y-2)^2=25
rendering as
{{{(x+5)^2+(y-2)^2=25}}}



Circle center is  (-5,2) and unknown point on the circle is {{{y=2+- sqrt(25-(x+5)^2)}}}.
The given point (2,8) with these other two points must form a right angle ON the circle
because, the point on the circle must be a tangent point, forming right angle with a radius.
Slope of the radius and slope of the tangency line must be negative reciprocals of each other.


Either the MINUS or the PLUS branch for the circle can be used.


{{{highlight_green(((2+sqrt(25-(x+5)^2)-2)/(x-(-5)))*((2+sqrt(25-(x+5)^2)-8)/(x-2))=-1)}}}
That is basically again, the product of the slopes must be negative ONE.


I leave the rest of the work and solution unfinished but for you to do.

(Several algebra steps done on paper seem to lead to  {{{17x^2+100x+20=0}}}, done in three-quarters of a page-worth).