What is the tangent of (x+2)^2+(y-3)^2=37,(-3,9) ?

The tangent of (x+2)^2+(y-3)^2=37,(-3,9) is y= 1/6 x+19/2

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tangent of (x+2)^2+(y-3)^2=37,(-3,9)

tangent of

(x + 2)^{2} + (y - 3)^{2} = 37, (- 3, 9)

y = \frac{1}{6} x + \frac{19}{2}

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

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

Find the line with slope m=

\frac{1}{6}

and passing through

(- 3, 9) : y = \frac{1}{6} x + \frac{19}{2}

y = \frac{1}{6} x + \frac{19}{2}

y = x^{2} + 9

y^{^{'}} + \frac{y}{t} = e^{t^{2}}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = \frac{e ^{x}}{1 + x ^{2}}

n \to \infty lim ({(1 + \frac{2}{n})^{4 n}})

y = x^{2} e^{x} - 2 x e^{x} + 9 e^{x}

What is the tangent of (x+2)^2+(y-3)^2=37,(-3,9) ?
The tangent of (x+2)^2+(y-3)^2=37,(-3,9) is y= 1/6 x+19/2