What is the tangent of (x+2)^2+(y-3)^2=37,(-8,2) ?

The tangent of (x+2)^2+(y-3)^2=37,(-8,2) is y=-6x-46

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tangent of (x+2)^2+(y-3)^2=37,(-8,2)

tangent of

(x + 2)^{2} + (y - 3)^{2} = 37, (- 8, 2)

y = - 6 x - 46

Solution steps

Find the slope of

(x + 2)^{2} + (y - 3)^{2} = 37 : \frac{d y}{d x} = - \frac{x + 2}{y - 3}

Compute the slope of

(x + 2)^{2} + (y - 3)^{2} = 37

(- 8, 2) : m = - 6

Find the line with slope m=

- 6

and passing through

(- 8, 2) : y = - 6 x - 46

y = - 6 x - 46

f (x) = x^{3} + 2 x, at x = - 1

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

f (x) = x ∣ x ∣

f (x) = 5 x - 4

\int_{\frac{1}{3}}^{5} 12 x ln (3 x) d x

What is the tangent of (x+2)^2+(y-3)^2=37,(-8,2) ?
The tangent of (x+2)^2+(y-3)^2=37,(-8,2) is y=-6x-46