# Tangent line to $$$- x^{2} + y^{2} - 4 y + 12 = 0$$$ at $$$\left(x, y\right) = \left(3, 1\right)$$$

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### Your Input

**Calculate the tangent line to $$$- x^{2} + y^{2} - 4 y + 12 = 0$$$ at $$$\left(x, y\right) = \left(3, 1\right)$$$.**

### Solution

We are given that $$$x_{0} = 3$$$, $$$y_{0} = 1$$$.

The slope of the tangent line at $$$\left(x, y\right) = \left(x_{0}, y_{0}\right)$$$ is the derivative of the function, evaluated at $$$\left(x, y\right) = \left(x_{0}, y_{0}\right)$$$: $$$M{\left(x_{0},y_{0} \right)} = \frac{dy}{dx}|_{\left(\left(x, y\right) = \left(x_{0}, y_{0}\right)\right)}$$$.

Differentiate implicitly: $$$\frac{dy}{dx} = \frac{x}{y - 2}$$$ (for steps, see implicit derivative calculator).

Hence, $$$M{\left(x_{0},y_{0} \right)} = \frac{dy}{dx}|_{\left(\left(x, y\right) = \left(x_{0}, y_{0}\right)\right)} = \frac{x_{0}}{y_{0} - 2}$$$.

Next, find the slope at the given point.

$$$m = M{\left(3,1 \right)} = -3$$$

Finally, the equation of the tangent line is $$$y - y_{0} = m \left(x - x_{0}\right)$$$.

Plugging the found values, we get that $$$y - 1 = - 3 \left(x - 3\right)$$$.

Or, more simply: $$$y = 10 - 3 x$$$.

### Answer

**The equation of the tangent line is $$$y = 10 - 3 x$$$A.**