The equation of a circle is given by D:
x^2 + y^2 - 2x - 4y - 20 = 0
Use implicit differentiation to find the equation of the normal D at the point S(5, -1)
find slope of tangent dy/dx
2 x + 2 y dy/dx - 2 -4 dy/dx = 0
dy/dx (2y-4) = 4 - 2x
dy/dx = (4-2x)/(2y-4)
at x = 5 and y = -1
dy/dx = (4 -10)/(-6) = -6/-6 = 1 :)
slope of normal = -1/slope
= -1
so
y = -x + b
-1 = -5 + b
b = 4
so in the end
y = -x + 4