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:

${(x-h)}^2+{(y-k)}^2=r^2$

Throwing our value for k in:

${(x-h)}^2+{(y-3)}^2=r^2$

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:

${(0-h)}^2+{(3-3)}^2=r^2$

$h^2=r^2$

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:

${(1-r)}^2+{(0-3)}^2=r^2$

$1-2r+r^2+9=r^2$

Through some algebraic manipulation:

$2r=10$

$r=h=5$

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.

${(x-5)}^2+{(0-3)}^2=5^2$

$x^2-10x+25+9=25$

$x^2-10x+9=0$

${(x-9)}{(x-1)}=0$

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.