# Slopes and Circles

Alignments to Content Standards: G-GPE.A.1

Suppose $A = (a_1, a_2)$ and $B = (b_1, b_2)$ are two points in the plane, determined by constants $a_1, a_2, b_1, b_2$. Suppose $X = (x_1, x_2)$ is a third point, determined by the variables $x_1$ and $x_2$.

1. Write an expression that gives the slope of line $AX$ in terms of $a_1, a_2, x_1, x_2$. Write an expression that gives the slope of line $BX$ in terms of $b_1, b_2, x_1, x_2$.

2. Write a polynomial equation involving $a_1, a_2, b_1, b_2, x_1, x_2$ that expresses that the lines $AX$ and $BX$ are perpendicular.

3. What geometric figure is the solution set of the equation in b)?

## IM Commentary

The purpose of this task is to lead students through an algebraic approach to a well-known result from classical geometry, namely, that a point $X$ is on the circle of diameter $AB$ whenever $\angle AXB$ is a right angle. This consists of translating the right-angle condition into the negative reciprocal slope property of perpendicular lines, followed by a short series of algebraic manipulations (principally, completing the square).

Teachers whose students are seeing this geometric result (which is sometimes referred to as the converse to Thales's Theorem) for the first time may wish to preface this task with a numerical exploration of the phenomena, e.g., having them find explicit points $X$ giving a right $\angle AXB$ for $A=(-1,0)$ and $B=(1,0)$. Teachers looking to reduce the algebraic difficulty could specify explicitly that part (c) can be done by completing the square.

## Solution

1. Letting $m_{AX}$ and $m_{BX}$ denote the slopes of line $AX$ and $BX$, respectively, we have: $$m_{AX} = \frac{a_2 - x_2}{a_1 - x_1} \qquad m_{BX} = \frac{b_2 - x_2}{b_1 - x_1}$$

2. Using the slope criterion for perpendicularity: \begin{align} AX \perp BX \Leftrightarrow & m_{AX} \cdot m_{BX} = -1 \\ \Leftrightarrow & \frac{a_2 - x_2}{a_1 - x_1} \cdot \frac{b_2 - x_2}{b_1 - x_1} = -1 \\ \Leftrightarrow & (a_2 - x_2)(b_2 - x_2) = (-1)(a_1 - x_1)(b_1 - x_1) \tag{1} \end{align}

3. Equation (1) is equivalent to: $$0=x_{1}^{2} âˆ’ (a_1 +b_1)x_1 +a_1b_1 +x_{2}^2 âˆ’(a_2 +b_2)x_2 +a_2b_2$$

After completing the squares and some simplifying, this equations transforms to: $$\left( \frac{a_1-b_1}{2} \right)^2 + \left( \frac{a_2-b_2}{2} \right)^2 = \left( x_1 - \left( \frac{a_1+b_1}{2} \right) \right) ^2 + \left( x_2 - \left( \frac{a_2+b_2}{2} \right) \right) ^2$$

By the distance formula (or by the Pythagorean Theorem), the left side is the square of half the length of segment $AB$. The constants $\frac{a_1+b_1}{2}$ and $\frac{a_2+b_2}{2}$ on the right are the coordinates of the midpoint of segment $AB$. Thus, the solution of the equation is the circle of radius $|AB|/2$, with center at the midpoint of segment $AB$.