# How do you find the equation of the normal to the circle x^2 + y^2 = 40 at the point (6,2)?

Jun 6, 2016

$y = \frac{1}{3} x$

#### Explanation:

Note that by circle properties, since the tangent is perpendicular to the radius of the circle at the point $\left(6 , 2\right)$, the normal, which is perpendicular to the tangent, must be parallel to the radius.

Thus, all we need is the gradient of the normal in order to find its equation, since we are given a fixed point $\left(6 , 2\right)$.

Gradient $= \frac{2 - 0}{6 - 0} = \frac{1}{3}$

Equation of normal is $y - 2 = \frac{1}{3} \left(x - 6\right)$ and thus $y = \frac{1}{3} x$.