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

The tangent of (x+2)^2+(y-3)^2=37,(-1,9) is y=-1/6 x+53/6

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tangent of (x+2)^2+(y-3)^2=37,(-1,9)

tangent of

(x + 2)^{2} + (y - 3)^{2} = 37, (- 1, 9)

y = - \frac{1}{6} x + \frac{53}{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, 9) : m = - \frac{1}{6}

Find the line with slope m=

- \frac{1}{6}

and passing through

(- 1, 9) : y = - \frac{1}{6} x + \frac{53}{6}

y = - \frac{1}{6} x + \frac{53}{6}

\int tan (4 x) sec^{2} (4 x) d x

x^{2} - e^{y^{2}}

y = arctan (e^{x})

\int 3 x^{2} - 6 x d x

f (x) = x^{2} + x - 56, g (x) = - x^{2} + 3 x + 4

What is the tangent of (x+2)^2+(y-3)^2=37,(-1,9) ?
The tangent of (x+2)^2+(y-3)^2=37,(-1,9) is y=-1/6 x+53/6