What is the tangent of (x-5)^2+(y-3)^2=18,(8,6) ?

The tangent of (x-5)^2+(y-3)^2=18,(8,6) is y=-x+14

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tangent of (x-5)^2+(y-3)^2=18,(8,6)

tangent of

(x - 5)^{2} + (y - 3)^{2} = 18, (8, 6)

y = - x + 14

Solution steps

Find the slope of

(x - 5)^{2} + (y - 3)^{2} = 18 : \frac{d y}{d x} = \frac{- x + 5}{- 3 + y}

Compute the slope of

(x - 5)^{2} + (y - 3)^{2} = 18

(8, 6) : m = - 1

Find the line with slope m=

- 1

and passing through

(8, 6) : y = - x + 14

y = - x + 14

\int_{0}^{4} π (3 x)^{2} d x

f (x) = cos (tan (x))

y^{^{''}} + a y^{^{'}} + b y = t + e^{3 t}

x \to - 1 lim (ln (x + 1))

\frac{\partial}{\partial x} (\frac{1}{2} (x^{2} + 4 y^{2}))

What is the tangent of (x-5)^2+(y-3)^2=18,(8,6) ?
The tangent of (x-5)^2+(y-3)^2=18,(8,6) is y=-x+14