SOLUTION: Find the length of the tangent line segment from (2,8) to (x+5)2+(y-2)2=25.

Algebra ->  Length-and-distance -> SOLUTION: Find the length of the tangent line segment from (2,8) to (x+5)2+(y-2)2=25.      Log On


   

Question 1018000: Find the length of the tangent line segment from (2,8) to (x+5)2+(y-2)2=25.
Answer by josgarithmetic(39631) About Me  (Show Source):
You can put this solution on YOUR website!
(x+5)2+(y-2)2=25
change to
(x+5)^2+(y-2)^2=25
rendering as
%28x%2B5%29%5E2%2B%28y-2%29%5E2=25


Circle center is (-5,2) and unknown point on the circle is y=2%2B-+sqrt%2825-%28x%2B5%29%5E2%29.
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.


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%5E2%2B100x%2B20=0, done in three-quarters of a page-worth).