# Given the circle (x – 6)^2 + (y + 2)^2 = 9, how do you determine if the line with an equation of x = 9 is a tangent to the circle, a secant to the circle, or neither?

May 22, 2016

When you're in the form ${\left(x - n\right)}^{2} + {\left(y - m\right)}^{2} = {r}^{2}$, the point $\left(n , m\right)$ is the center of the circle and r is the radius.

#### Explanation:

Knowing this information, we can deduce that the center of the circle is at $\left(6 , - 2\right)$, and the radius measures 3 units.

$x = 9$ is a vertical line, while the circle will extend equal length in all direction, in function of it's radius. Looking at the centre and adding 3 to the x value, we get the point $\left(9 , - 2\right)$. This point doesn't only lie on the circle; it also lies on the line $x = 9$. Furthermore, this is the furthest possible distance from the center, so the line will only pass through this one point. Therefore, we can state that the line x = 9 is tangent to ${\left(x - 6\right)}^{2} + {\left(y + 2\right)}^{2} = 9$.

If you graph the circle and the line you will get the same answer.

Hopefully this helps!