What is the tangent of (x+2)^2+(y-3)^2=37,(4,4) ?

The tangent of (x+2)^2+(y-3)^2=37,(4,4) is y=-6x+28

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tangent of (x+2)^2+(y-3)^2=37,(4,4)

tangent of

(x + 2)^{2} + (y - 3)^{2} = 37, (4, 4)

y = - 6 x + 28

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

(4, 4) : m = - 6

Find the line with slope m=

- 6

and passing through

(4, 4) : y = - 6 x + 28

y = - 6 x + 28

\int \frac{x}{1} d x

x = 5 y^{2}, x = 1 + 4 y^{2}

3 x^{- \frac{3}{2}} + 6 x^{- \frac{1}{2}} + x^{3} - 4

y = e^{3 x}

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

What is the tangent of (x+2)^2+(y-3)^2=37,(4,4) ?
The tangent of (x+2)^2+(y-3)^2=37,(4,4) is y=-6x+28