Question 1088210
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.