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

The tangent of (x+2)^2+(y-3)^2=37,(-3,-3) is y=-1/6 x-7/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

tangent of

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

y = - \frac{1}{6} x - \frac{7}{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, - 3) : m = - \frac{1}{6}

Find the line with slope m=

- \frac{1}{6}

and passing through

(- 3, - 3) : y = - \frac{1}{6} x - \frac{7}{2}

y = - \frac{1}{6} x - \frac{7}{2}

\int_{0}^{\frac{π}{4}} 3 cos^{5} (2 x) d x

\int - 2 t cos (t) d t

\int 48 t^{7} e^{- t^{8}} d t

\int x (3 x + 10)^{49} d x

\frac{d}{d x} ((x + 39)^{2} + (43 + x)^{2})

What is the tangent of (x+2)^2+(y-3)^2=37,(-3,-3) ?
The tangent of (x+2)^2+(y-3)^2=37,(-3,-3) is y=-1/6 x-7/2