Question 1106128
The problem here is that the point (11, 5) is NOT on the circle but (sqrt (11), 5) is
x^2+y^2=36
implicit differentiation
2x+2y dy/dx=0
2y dy/dx=-2x
dy/dx=-x/y, and that is the slope of the tangent line to the circle.
calling it m, m=-sqrt(11)/5

point slope formula y-y1=m(x-x1), m slope, (x1, y1) point
y-5=(-sqrt(11)/5)(x-sqrt (11))
y-5=-(sqrt(11)/5)x+(11/5)
y=-(sqrt(11)/5)x+7.2

{{{graph(300,300,-10,10,-10,10,sqrt(36-x^2),-sqrt(36-x^2),-0.66x+7.2)}}}