Circling a Solution

One of my old students emailed me about her summer work heading into calculus–

A circle is tangent To the y-axis at y=3 and has one x-intercept at x =1. Determine the other x-intercept.

What I enjoy about the problem, is that the harder parts of it are more deeply related to finding a way to analyze the problem, then the actual math involved here.

First:  If we know the circle is tangent to the y-axis, we know that a radius to this point from the center will be perpendicular to the y-axis (i.e. horizontal), which is really useful:  it gives us a y-coordinate for the center to use in our general equation for a circle.

Let’s take a look at the algebra now:

The general equation for a circle given center (h,k) and radius r, is:


Throwing our value for k in:


We also happen to know a couple points on the circle:  (1,0) and (0,3).

Let’s plug in (0,3), and we arrive at:



So, h=r, which makes sense, because if we are starting from the y-axis (where we’re sitting at the point of tangency), the distance to the center, is going to be equal to the difference in the x-coordinates, or (h-0), or h.  Cool.

Let’s plug in (1,0), and r for h, and we arrive at:



Through some algebraic manipulation:



Once we have a value for the x-coordinate of the center of the circle, we can think about this graphically or algebraically.

Graphically, we have a circle with center (5,3), and an x-intercept at (1,0).  We can reflect the point over the vertical diameter, x=5, and get the point (9,0), which is the other x-intercept.

Algebraically, we can plug in y=0 to the final equation of our circle, and solve for x.





x = 1, and x = 9.

This is a really hard problem for my students to do, but not one that’s impossible—and when a student can make the first leap or second leap, they get excited and build resiliency.