# How do you find the equation of the line tangent to the graph x^2 + y^2 = 12 at point (sqrt3, 3)?

Jun 26, 2015

I assume that it is the slope, rather than the equation, you need help with.

#### Explanation:

Implicit Differentiation

Because this question was posted under Implicit Differentiation, we'll do it that way.

We want $\frac{\mathrm{dy}}{\mathrm{dx}}$ given that $x$ and $y$ are related by:

${x}^{2} + {y}^{2} = 12$

Differentiate both sides with respect to $x$. Remember that we are considering $y$ to be some function of $x$ that we haven't found. It often helps to think of $y$ as "some stuff in parentheses".

When we differentiate "some stuff in parentheses" squared, we gat $2$ times that stuff times the derivative of the stuff (because of the chain rule).
Also: remember that $\frac{d}{\mathrm{dx}}$ of some expression means the derivative (with respect to x) of that expression.

o, (in moderate detail)

${x}^{2} + {y}^{2} = 12$

$\frac{d}{\mathrm{dx}} \left({x}^{2} + {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(12\right)$

$\frac{d}{\mathrm{dx}} \left({x}^{2}\right) + \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(12\right)$

$2 x + {\underbrace{2 y \frac{\mathrm{dy}}{\mathrm{dx}}}}_{\text{used chain rule}} = 0$

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

At $\left(\sqrt{3} , 3\right)$, the slope is $m = - \frac{\sqrt{3}}{3}$

The equation of the line through that point with that slope is:

$y = - \frac{\sqrt{3}}{3} x + 4$

Notes

(1) because we see that the $y$ coordinate of the point of interest is positive, we know that the point lies on the upper half of the circle:

${x}^{2} + {y}^{2} = 12$

Solving for $y$ on that semicircle gets us $y = \sqrt{12 - {x}^{2}}$ and we do not need implicit differentiation.

(2) It is a circle centered at the origin, The tangent at $\left(\sqrt{3} , 3\right)$ is perpendicular to the radius connecting $\left(0 , 0\right)$ and $\left(\sqrt{3} , 3\right)$. Find the slope of the radius and then the slope of a perpendicular to the radius. No need for calculus.