# The Circle and The Line

Alignments to Content Standards: A-REI.C.7

Sketch the circle with equation $x^2 + y^2 = 1$ and the line with equation $y = 2x -1$ on the same pair of axes.

1. There is one solution to the pair of equations \begin{align} x^2 + y^2 &= 1\\ 2x-1 &= y \end{align} that is clearly identifiable from the sketch. What is it? Verify that it is a solution.

2. Find all the solutions to this pair of equations.

## IM Commentary

Although this task is fairly straightforward, it is worth noticing that it does not explicitly tell students to look for intersection points when they graph the circle and the line. Thus, in addition to assessing whether they can solve the system of equations, it is assessing a simple but important piece of conceptual understanding, namely the correspondence between intersection points of the two graphs and solutions of the system.

## Solution

The equations $x^2 + y^2 = 1$ and $y = 2x - 1$ are graphed below.

The solution that is clearly identifiable from the graph, the point at which the circle and line intersect, is $(0,-1)$.

We can check that $(0,-1)$ is a solution to the pair of equations by substituting (plugging in) 0 for $x$ and -1 for $y$ in both equations.

\begin{align} x^2 + y^2 &= 1 \\ (0)^2 + (-1)^2 &= 1 \\ 0 + 1 &= 1 \\ 1 &= 1 \quad \checkmark \\ \\ y &= 2x - 1 \\ -1 &= 2(0) - 1 \\ -1 &= 0 - 1 \\ -1 &= -1 \quad \checkmark \end{align}

We have verified that $(0,-1)$ is a solution to the pair of equations.

From the graph, we can see that there is another solution (in Quadrant I). However, it is difficult to visually determine its exact $x$- and $y$-coordinates. To find its exact location we can solve the system of equations by substitution.

Let $(x,y)$ be the intersection point. Since $y=2x-1$ by virtue of the point being on the line, we can substitute the quantity $(2x-1)$ for every $y$ appearing in the equation of the circle. We then simplify as follows: \begin{align} &x^2 + (2x-1)^2 = 1 \\ &x^2 + (2x-1)(2x-1) = 1 \\ &x^2 + 4x^2 - 2x - 2x + 1 = 1 \\ &5x^2 - 4x + 1 = 1 \\ &5x^2 - 4x = 0 \\ &x(5x - 4) = 0 \end{align} \begin{align} x = 0 \qquad \text{or} \qquad 5x - 4 &= 0 \\ 5x &= 4 \\ x &= \frac{4}{5} \end{align}

If $x=0$, we know $y=-1$, so we have re-discovered the first intersection point we observed. So our second intersection point has $x$-coordinate equal to $\frac{4}{5}$, and we are left only having to now find its $y$-coordinate. We simply substitute $x=\frac45$ in either equation and solve for $y$.

\begin{align} y &= 2x - 1 \ y &= 2\left(\frac45\right)-1 \ y &= \frac85 - 1 \ y &= \frac85 - \frac55 \ y &= \frac35 \end{align}

Now we have that $\left( \frac45, \frac35 \right)$ is also a solution.