# How do you find the equations for the normal line to x^2+y^2=9 through (0,3)?

Nov 29, 2016

$x = 0$

#### Explanation:

We find the derivative.

$2 x + 2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$2 y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = - 2 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 x}{2 y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$

We now determine the slope of the tangent.

${m}_{\text{tangent}} = - \frac{0}{3}$

${m}_{\text{tangent}} = 0$

The normal line is perpendicular to the tangent, so the slope of the normal line will be $- \frac{1}{k}$ of that of the tangent, where $k$ is the slope of the tangent.

${m}_{\text{normal}} = \emptyset$

Hence, the normal line will be vertical at $\left(0 , 3\right)$ and will be of the form $x = a$.

Since the point of intersection is $x = 0$, the normal line will have equation $x = 0$.

Hopefully this helps!

Nov 29, 2016

Here is a solution using geometry.

#### Explanation:

${x}^{2} + {y}^{2} = 9$ is a circle centered at $\left(0 , 0\right)$ with radius $3$.

The point $\left(0 , 3\right)$ is on the $y$-axis.

The line segment joining $\left(0 , 0\right)$ and $\left(0 , 3\right)$ is a radius of the circle and the radius is perpendicular to the tangent.

So the $y$ axis is the normal line at that point.

Nov 29, 2016

Here is a solution using calculus, but without implicit differentiation.

#### Explanation:

${x}^{2} + {y}^{2} = 9$

$y = \pm \sqrt{9 - {x}^{2}}$

The point $\left(0 , 3\right)$ is a solution to $y = \sqrt{9 - {x}^{2}}$ so we are interested in the function $f \left(x\right) = \sqrt{9 - {x}^{2}}$.

Differentiate using the chain rule to get

$f ' \left(x\right) = \frac{1}{2 \sqrt{9 - {x}^{2}}} \cdot \left(- 2 x\right) = \frac{- x}{\sqrt{9 - {x}^{2}}}$

At $\left(0 , 3\right)$, the tangent line has slope $f ' \left(0\right) = 0$. So the tangent line is a horizontal line and the normal line is vertical.

The vertical line through $\left(0 , 3\right)$ has equation $x = 0$.