Question

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?

Answer 1

When you're in the form `(x - n)^2 + (y - m)^2 = r^2`, the point `(n, m)` 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 `(6, -2)`, 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 `(9, -2)`. 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 `(x - 6)^2 + (y + 2)^2 = 9`.

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

Hopefully this helps!