SOLUTION: Find the equations of the lines described. Give your answers in the form y=mx+b. tangent lines to x�+y�-22x+12y+137=0 at points where x=15

Question 1088210: Find the equations of the lines described. Give your answers in the form y=mx+b. tangent lines to x�+y�-22x+12y+137=0 at points where x=15
Found 2 solutions by natolino_2017, Fombitz:
Answer by natolino_2017(77) (Show Source): You can put this solution on YOUR website!
first we need dy/dx, by using implicit derivative over the relation
2x + 2y(dy/dx) - 22 + 12dy/dx = 0
dy/dx = (11-x)/(y+6) for y different to -6.
then, we need to find the point(s) when x = 15 using the relation:
(15)^2 + y^2 - 22(15) + 12y + 137 = 0
Solving for y: y^2 + 12y + 32 = 0
has two solutions y = {-4, -8} So the point are P =(15,-4) and Q= (15,-8).
first when the point is P the dy/dx = (11-15)/(-4+6) = -2 (slope)
so the line is L : y = -2x +b, knowing that P belongs to the line.
-4 = -2(15) + b, so b = 26.
Second for the point Q, dy/dx = (11-15)/(-8+6) = 2 (slope)
so the line is L2: y = 2x + c, knowing that Q belongs to the line.
-8 = 2(15) +c, so c = -38.
Answer: L: y = -2x +26 , L2: y = 2x -38, which are tangent lines to the intersection of the curve with the line x = 15.

Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
The slope of the tangent line is equal to the value of the derivative at the point.
Complete the square to get to center-radius form of a circle,

So implicitly differentiating,

So at , find the y value,

So then,

and

So then using the point-slope form of the line,

and

I leave it to you to convert to slope-intercept form.
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