What is the tangent of x^2+y^2=25,(4,3) ?

The tangent of x^2+y^2=25,(4,3) is y=-4/3 x+25/3

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tangent of x^2+y^2=25,(4,3)

tangent of

x^{2} + y^{2} = 25, (4, 3)

y = - \frac{4}{3} x + \frac{25}{3}

Solution steps

Find the slope of

x^{2} + y^{2} = 25 : \frac{d y}{d x} = - \frac{x}{y}

Compute the slope of

x^{2} + y^{2} = 25

(4, 3) : m = - \frac{4}{3}

Find the line with slope m=

- \frac{4}{3}

and passing through

(4, 3) : y = - \frac{4}{3} x + \frac{25}{3}

y = - \frac{4}{3} x + \frac{25}{3}

\int - \frac{2}{x ^{2} + 1} d x

\frac{\partial}{\partial x} (5 e^{x y + 4})

y^{^{''}} + 25 y = 2 tan (5 x)

\frac{2 s + 5}{s ^{2} + 5 s + 6}

\frac{d}{d x} (arctan (g (x)) x)

What is the tangent of x^2+y^2=25,(4,3) ?
The tangent of x^2+y^2=25,(4,3) is y=-4/3 x+25/3