Lesson 11.1: Distance and Midpoint Formulas; Circles

New Concept

We'll use the Pythagorean theorem to derive the Distance and Midpoint formulas. These tools allow us to connect the geometry of a circle (its center and radius) to its standard algebraic equation, $(x-h)^2 + (y-k)^2 = r^2$.

What’s next

Now, put these formulas into practice! You'll work through interactive examples, watch short videos, and solve problems to master finding distances, midpoints, and graphing circles.

Property

The distance $d$ between the two points $(x_1, y_1)$ and $(x_2, y_2)$ is

Examples

• To find the distance between $(1, 2)$ and $(4, 6)$, use the formula: $d = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
• The distance between $(-2, 5)$ and $(10, 0)$ is calculated as $d = \sqrt{(10 - (-2))^2 + (0 - 5)^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$.
• Find the distance between $(2, 5)$ and $(4, -1)$. The exact distance is $d = \sqrt{(4 - 2)^2 + (-1 - 5)^2} = \sqrt{2^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40}$. This is approximately $6.3$.

Explanation

This formula is the Pythagorean theorem applied to the coordinate plane. It calculates the straight-line distance between two points by treating that distance as the hypotenuse of a right triangle formed by the horizontal and vertical differences.

Property

The midpoint of the line segment whose endpoints are the two points $(x_1, y_1)$ and $(x_2, y_2)$ is

To find the midpoint of a line segment, we find the average of the $x$-coordinates and the average of the $y$-coordinates of the endpoints.

Examples

• The midpoint of a segment with endpoints $(2, 5)$ and $(8, 11)$ is found by averaging the coordinates: $\left( \frac{2 + 8}{2}, \frac{5 + 11}{2} \right) = \left( \frac{10}{2}, \frac{16}{2} \right) = (5, 8)$.
• For endpoints $(-4, 9)$ and $(6, -3)$, the midpoint is $\left( \frac{-4 + 6}{2}, \frac{9 + (-3)}{2} \right) = \left( \frac{2}{2}, \frac{6}{2} \right) = (1, 3)$.
• The midpoint of the segment between $(3, 8)$ and $(10, 5)$ is $\left( \frac{3 + 10}{2}, \frac{8 + 5}{2} \right) = \left( \frac{13}{2}, \frac{13}{2} \right)$.

Explanation

To find the exact center of a line segment, you simply calculate the average of the x-coordinates and the average of the y-coordinates. This gives you the coordinates of the point that is perfectly in the middle.

Property

The standard form of the equation of a circle with center, $(h, k)$, and radius, $r$, is

Examples

• The equation of a circle with radius $5$ and center $(0, 0)$ is $x^2 + y^2 = 5^2$, which simplifies to $x^2 + y^2 = 25$.
• A circle with radius $4$ and center $(3, -2)$ has the equation $(x - 3)^2 + (y - (-2))^2 = 4^2$, which simplifies to $(x - 3)^2 + (y + 2)^2 = 16$.
• For a circle with center $(1, 2)$ that passes through $(4, 6)$, first find the radius using the distance formula: $r = \sqrt{(4-1)^2 + (6-2)^2} = 5$. The equation is $(x - 1)^2 + (y - 2)^2 = 25$.

Explanation

This equation describes every point $(x, y)$ that is exactly distance $r$ from the center $(h, k)$. It comes directly from the Distance Formula, where the distance is the radius, and both sides are squared.

Property

The general form of the equation of a circle is

To find the center and radius from this form, group the x-terms and y-terms, move the constant to the right side, and complete the square for both variables to rewrite the equation in standard form.

Examples

• Given $x^2 + y^2 - 8x + 2y + 8 = 0$, complete the square: $(x^2 - 8x + 16) + (y^2 + 2y + 1) = -8 + 16 + 1$. The standard form is $(x - 4)^2 + (y + 1)^2 = 9$. The center is $(4, -1)$ and the radius is $3$.
• Given $x^2 + y^2 + 10y - 11 = 0$, complete the square for y: $x^2 + (y^2 + 10y + 25) = 11 + 25$. The standard form is $x^2 + (y + 5)^2 = 36$. The center is $(0, -5)$ and the radius is $6$.
• Given $x^2 + y^2 - 6x + 5 = 0$, complete the square for x: $(x^2 - 6x + 9) + y^2 = -5 + 9$. The standard form is $(x - 3)^2 + y^2 = 4$. The center is $(3, 0)$ and the radius is $2$.

Explanation

The general form is a 'scrambled' version of the standard circle equation. By completing the square, you can unscramble it back into standard form, which clearly reveals the circle's center and radius, making it easy to graph.