What is the tangent of 5=(x+9)^2+(y-2)^2,(-7,3) ?

The tangent of 5=(x+9)^2+(y-2)^2,(-7,3) is y=-2x-11

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tangent of 5=(x+9)^2+(y-2)^2,(-7,3)

tangent of

5 = (x + 9)^{2} + (y - 2)^{2}, (- 7, 3)

y = - 2 x - 11

Solution steps

Find the slope of

5 = (x + 9)^{2} + (y - 2)^{2} : \frac{d y}{d x} = - \frac{x + 9}{y - 2}

Compute the slope of

5 = (x + 9)^{2} + (y - 2)^{2}

(- 7, 3) : m = - 2

Find the line with slope m=

- 2

and passing through

(- 7, 3) : y = - 2 x - 11

y = - 2 x - 11

e^{2 x} \frac{d y}{d x} - e^{x} y = e^{- x} y^{2}

\frac{d}{d x} (\frac{- x}{1 + x ^{2}})

y = x^{3} - x^{2}, y = 0

\frac{\partial}{\partial x} (3 x^{2} y^{4} + 1)

\int \frac{1}{x 9 x ^{2} - 4} d x

What is the tangent of 5=(x+9)^2+(y-2)^2,(-7,3) ?
The tangent of 5=(x+9)^2+(y-2)^2,(-7,3) is y=-2x-11