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?

1 Answer
May 22, 2016

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.


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!