What is the tangent of (x+2)^2+(y-3)^2=37,(-1,-3) ?

The tangent of (x+2)^2+(y-3)^2=37,(-1,-3) is y= 1/6 x-17/6

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tangent of (x+2)^2+(y-3)^2=37,(-1,-3)

tangent of

(x + 2)^{2} + (y - 3)^{2} = 37, (- 1, - 3)

y = \frac{1}{6} x - \frac{17}{6}

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

(- 1, - 3) : m = \frac{1}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(- 1, - 3) : y = \frac{1}{6} x - \frac{17}{6}

y = \frac{1}{6} x - \frac{17}{6}

\frac{d}{d x} (\frac{x ^{2} - 2}{x ^{3} + 3 x ^{2}})

y = \frac{x + x ^{3}}{x ^{2}}

f (x) = 7 - 2 x

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

\int 2 \cdot x \cdot ln (4 x) d x

What is the tangent of (x+2)^2+(y-3)^2=37,(-1,-3) ?
The tangent of (x+2)^2+(y-3)^2=37,(-1,-3) is y= 1/6 x-17/6