How to find slope of tangent line to the circle at pt. A (5, 14)?

Question

Circle equation: (x-17)^2 + (y-19)^2 = 169

Answer 1

⇒(x-17)² + (y-19)² = 169

Slope is y ' = dy/dx = 0

Take a derivative with respect to 'x'

⇒(d/dx)(x-17)² + (d/dx)(y-19)² = (d/dx)169

Note: (d/dx)xⁿ = nx^(n-1) and (d/dx)(constant) = 0

⇒ 2(x-17) + 2(y-19)(dy/dx) = 0

Subtract '2(x-17)' from each side.

⇒ 2(x-17) + 2(y-19)(dy/dx) - 2(x-17) = -2(x-17)

Simplify

⇒ 2(y-19)(dy/dx) = -2(x-17)

Divide each side by '2(y-19)'

⇒ [2(y-19)(dy/dx)] / [2(y-19)] = [-2(x-17)] / [2(y-19)]

Simplify

⇒ (dy/dx) = - (x-17) / (y-19)

Substitute in your values (5, 14) for the slope

⇒ dy/dx = - (5 -17 ) / ( 14 - 19 )

⇒ dy/dx = - (-12) / (-5)

⇒ dy/dx = - 12 / 5

slope = - 12/5