00.09.02 · precalc / coordinate-geom

Coordinate geometry — distance, lines, and circles

shipped3 tiersLean: none

Anchor (Master): Descartes 1637 La Géométrie (Leiden) Book II; Fermat 1636 Ad Locos Planos et Solidos Isagoge.

Intuition Beginner

The coordinate plane turns geometry into algebra. Every point becomes a pair of numbers $(x, y)$ . A straight line becomes an equation you can write down and solve. A circle becomes an equation too. Shape questions — how far apart, do they meet, are they parallel — become number questions you handle with arithmetic.

Distance comes from a right triangle. The horizontal gap between two points is one leg and the vertical gap is the other. The straight-line distance is the hypotenuse, so Pythagoras gives it. A line's steepness is its slope, the ratio of rise to run. Two lines are parallel when their slopes match and perpendicular when their slopes multiply to $- 1$ .

A circle is every point at a fixed distance (the radius) from a center. Writing that condition as an equation produces the circle form. Once lines and circles both have equations, you can answer geometric questions by algebra: solving the equations together tells you whether and where the shapes meet.

Visual Beginner

Object	Geometric description	Equation
Distance	hypotenuse of the gap right triangle	$d = (Δ x)^{2} + (Δ y)^{2}$
Line	constant slope between any two of its points	$y = m x + b$
Perpendicular lines	slopes multiply to $- 1$	$m_{1} m_{2} = - 1$
Circle	all points at distance $r$ from $(h, k)$	$(x - h)^{2} + (y - k)^{2} = r^{2}$

The pattern is the same throughout: a geometric condition (fixed distance, fixed slope, fixed direction) becomes an algebraic equation. Reading the equation tells you the shape and reading the shape tells you the equation.

Worked example Beginner

Example 1 — distance. Find the distance between $A = (1, 2)$ and $B = (4, 6)$ .

Horizontal gap: $4 - 1 = 3$ . Vertical gap: $6 - 2 = 4$ . These are the legs. The distance is the hypotenuse:

3^{2} + 4^{2} = 9 + 16 = 25 = 5.

Example 2 — a line. Find the equation of the line through $(1, 2)$ and $(3, 8)$ .

Slope is rise over run: $m = (8 - 2) / (3 - 1) = 6/2 = 3$ . Using point-slope through $(1, 2)$ : $y - 2 = 3 (x - 1)$ . Solving for $y$ gives $y = 3 x - 1$ . The slope is $3$ and the $y$ -intercept is $- 1$ .

Example 3 — a circle. Write the equation of the circle centered at $(2, - 1)$ with radius $3$ .

Every point $(x, y)$ on the circle is distance $3$ from $(2, - 1)$ . Squaring the distance condition gives $(x - 2)^{2} + (y + 1)^{2} = 9$ .

Check your understanding Beginner

Formal definition Intermediate+

Throughout, $P_{1} = (x_{1}, y_{1})$ and $P_{2} = (x_{2}, y_{2})$ are points in $R^{2}$ ^{[Stewart — Precalculus Ch. 1]}.

Distance and midpoint. The straight-line distance is

d (P_{1}, P_{2}) = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2},

and the midpoint of the segment $P_{1} P_{2}$ is

M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}) .

Slope. For $x_{1} \neq = x_{2}$ the slope of the line through $P_{1}$ and $P_{2}$ is $m = (y_{2} - y_{1}) / (x_{2} - x_{1})$ . A vertical line has undefined slope.

Line equations — four interchangeable forms.

Slope-intercept: $y = m x + b$ , with slope $m$ and $y$ -intercept $b$ .
Point-slope: $y - y_{0} = m (x - x_{0})$ , through $(x_{0}, y_{0})$ with slope $m$ .
Two-point: $y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})$ for $x_{1} \neq = x_{2}$ .
General: $A x + B y + C = 0$ with $(A, B) \neq = (0, 0)$ . When $B \neq = 0$ this is $y = (- A / B) x + (- C / B)$ .

Converting between forms is mechanical: expand point-slope to read off slope and intercept, or collect terms of $y = m x + b$ into $A x + B y + C = 0$ . The general form uniquely handles vertical lines ( $B = 0$ ), which the slope-intercept form cannot write.

Parallel and perpendicular. Two non-vertical lines with slopes $m_{1}, m_{2}$ are parallel iff $m_{1} = m_{2}$ and perpendicular iff $m_{1} m_{2} = - 1$ . A vertical line is perpendicular to a horizontal line.

Circle as a locus. The circle with center $C = (h, k)$ and radius $r > 0$ is the set of points at distance $r$ from $C$ :

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

Expanding gives the general second-degree circle equation $x^{2} + y^{2} + D x + E y + F = 0$ (the coefficients of $x^{2}$ and $y^{2}$ are equal and there is no $x y$ term). Completing the square recovers the center and radius:

x^{2} + y^{2} + D x + E y + F = 0 ⟺ (x + \frac{D}{2})^{2} + (y + \frac{E}{2})^{2} = \frac{D ^{2} + E ^{2}}{4} - F .

The equation represents a genuine circle iff $\frac{D ^{2} + E ^{2}}{4} - F > 0$ , a single point iff it equals $0$ , and the empty set iff it is negative.

Converting between descriptions and equations

Three points $\to$ circle: substituting each point into $x^{2} + y^{2} + D x + E y + F = 0$ yields three linear equations in $D, E, F$ .
Center and tangent line $\to$ circle: if a line is tangent to the circle, the distance from the center to the line equals the radius.
Diameter $\to$ circle: for endpoints $A, B$ , the center is the midpoint of $A B$ and the radius is $d (A, B) /2$ .

Key theorem with proof [Intermediate+] {#key-theorem}

Theorem (distance formula, Pythagorean origin). For $P_{1} = (x_{1}, y_{1})$ and $P_{2} = (x_{2}, y_{2})$ in $R^{2}$ ,

d (P_{1}, P_{2}) = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} .

Proof. Form the right triangle with vertices $P_{1}$ , $P_{2}$ , and the corner $Q = (x_{2}, y_{1})$ whose sides run parallel to the axes. The leg $P_{1} Q$ is horizontal with length $∣ x_{2} - x_{1} ∣$ and the leg $Q P_{2}$ is vertical with length $∣ y_{2} - y_{1} ∣$ . The segment $P_{1} P_{2}$ is the hypotenuse. By the Pythagorean theorem,

d (P_{1}, P_{2})^{2} = ∣ x_{2} - x_{1} ∣^{2} + ∣ y_{2} - y_{1} ∣^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2},

since $∣ a ∣^{2} = a^{2}$ for every real $a$ . Taking the non-negative square root gives the distance formula. $□$

Corollary (circle equation). The locus of points at distance $r$ from $(h, k)$ is $(x - h)^{2} + (y - k)^{2} = r^{2}$ , because substituting the distance formula into $d ((x, y), (h, k)) = r$ and squaring produces exactly this equation.

Bridge. The distance formula builds toward the whole circle-and-line machinery of this unit, and the same Pythagorean sum-of-squares appears again in 00.06.01 as the identity underlying right-triangle trigonometry and in 00.10.01 as the metric that defines conic sections by focus-directrix conditions. This is exactly the construction that promotes one-dimensional absolute-value distance 00.03.02 into the Euclidean metric on the plane, and the central insight is that squaring the distance kills the absolute values and turns a geometric locus into a polynomial equation. The bridge is between measurement (geometry) and equations (algebra): every planar locus becomes an algebraic condition on coordinate pairs, and the pattern generalises to the distance on $R^{n}$ and to the complex modulus on $C$ in 00.14.01. Putting these together, the Pythagorean theorem is the foundational reason coordinate geometry works at all.

Exercises Intermediate+

Exercise 8 (hard, symbolic).

Prove that the perpendicular bisectors of the three sides of a triangle meet at a single point (the circumcenter), and that this point is the center of the circle through the three vertices.

Hint

A perpendicular bisector is the locus of points equidistant from two vertices. Write the two bisector equations and observe that their intersection is equidistant from all three vertices.

Answer

Let the vertices be $A, B, C$ . The perpendicular bisector of $A B$ is ${(x, y) : d ((x, y), A) = d ((x, y), B)}$ , which squaring becomes a linear equation in $x$ and $y$ . Let $O$ be the intersection of the bisectors of $A B$ and $A C$ . Then $d (O, A) = d (O, B)$ and $d (O, A) = d (O, C)$ , so $d (O, A) = d (O, B) = d (O, C) =: r$ . The equality $d (O, B) = d (O, C)$ means $O$ also lies on the bisector of $B C$ , so all three bisectors meet at $O$ . The circle of radius $r$ centered at $O$ passes through $A, B, C$ .

Rubric: full credit for recognising the equidistance characterization, the linear-equation step, and the three-way equality argument.

Advanced results Master

The intermediate results specialise a coordinate framework in which lines and circles are solution sets of polynomial equations of degree one and two. The master-tier results handle the general second-degree circle, the distance from a point to a line, tangency, and the radical axis of two circles.

Theorem 1 (general second-degree circle). An equation $x^{2} + y^{2} + D x + E y + F = 0$ represents a circle with center $(- D /2, - E /2)$ and radius $(D^{2} + E^{2}) /4 - F$ when $(D^{2} + E^{2}) /4 - F > 0$ ; a single point when it is $0$ ; and the empty set when it is negative. The proof is completing the square in $x$ and in $y$ .

Theorem 2 (distance from a point to a line). The distance from $(x_{0}, y_{0})$ to the line $A x + B y + C = 0$ is

d = \frac{∣ A x _{0} + B y _{0} + C ∣}{A ^{2} + B ^{2}} .

The formula is invariant under scaling $(A, B, C)$ by a nonzero constant, so it depends only on the line.

Theorem 3 (tangency criterion). A line $L$ is tangent to the circle with center $C$ and radius $r$ iff $d (C, L) = r$ . Equivalently the system of the line and circle equations has exactly one solution. The proof inserts the parameterisation of $L$ into the circle equation and observes that the resulting quadratic has a repeated root exactly when its discriminant vanishes, which is the same condition as $d (C, L) = r$ .

Theorem 4 (radical axis). Given two non-concentric circles $S_{1} : x^{2} + y^{2} + D_{1} x + E_{1} y + F_{1} = 0$ and $S_{2} : x^{2} + y^{2} + D_{2} x + E_{2} y + F_{2} = 0$ , the set of points with equal power with respect to both circles is the line $(D_{1} - D_{2}) x + (E_{1} - E_{2}) y + (F_{1} - F_{2}) = 0$ . Subtracting the equations cancels the quadratic terms, leaving a linear locus.

Theorem 5 (reflection across a line). The reflection of $(x_{0}, y_{0})$ across $A x + B y + C = 0$ is

(x_{0}, y_{0}) - \frac{2 ( A x _{0} + B y _{0} + C )}{A ^{2} + B ^{2}} (A, B) .

Reflections preserve distance (they are orthogonal transformations), so the circle and its reflected image have equal radii.

Synthesis. The distance formula is the foundational reason every geometric locus in this unit admits an algebraic equation, and this is exactly the identification Descartes made between curves and equations. The bridge is between the metric structure of the plane and the algebra of polynomials: lines are degree-one loci, circles are degree-two loci, and tangency is the algebraic condition that a line and a circle share a single point. The general-circle formula generalises to arbitrary conic sections in 00.10.01, where completing the square becomes classifying a general second-degree curve, and the point-to-line distance generalises to the projection onto an affine subspace. Putting these together, the perpendicular-slope, tangency, and radical-axis results are all instances of one pattern: a geometric relation among points becomes the vanishing or equality of an algebraic expression, and the central insight is that the coordinates make the translation automatic.

Full proof set Master

Proposition 1 (perpendicular slope condition). Two non-vertical lines with slopes $m_{1}$ and $m_{2}$ are perpendicular iff $m_{1} m_{2} = - 1$ .

Proof. A line of slope $m$ has direction vector $(1, m)$ . For two lines through a common point with slopes $m_{1}, m_{2}$ the direction vectors are $v_{1} = (1, m_{1})$ and $v_{2} = (1, m_{2})$ , with dot product

v_{1} \cdot v_{2} = 1 + m_{1} m_{2} .

Two vectors are perpendicular iff their dot product vanishes, giving $1 + m_{1} m_{2} = 0$ , i.e. $m_{1} m_{2} = - 1$ . Translation does not change slope, so the same condition holds for lines not through a common point. $□$

Proposition 2 (general circle via completing the square). For real $D, E, F$ ,

x^{2} + y^{2} + D x + E y + F = 0 ⟺ (x + \frac{D}{2})^{2} + (y + \frac{E}{2})^{2} = \frac{D ^{2} + E ^{2}}{4} - F .

Proof. Add $D^{2} /4$ and $E^{2} /4$ to both sides of $x^{2} + D x + y^{2} + E y = - F$ . The left becomes a sum of two perfect squares:

(x^{2} + D x + \frac{D ^{2}}{4}) + (y^{2} + E y + \frac{E ^{2}}{4}) = \frac{D ^{2}}{4} + \frac{E ^{2}}{4} - F,

i.e. $(x + D /2)^{2} + (y + E /2)^{2} = (D^{2} + E^{2}) /4 - F$ . Reading off the radius requires the right-hand side to be positive; it is $0$ for a point-circle and negative for the empty set. $□$

Proposition 3 (distance from a point to a line). The distance from $(x_{0}, y_{0})$ to $A x + B y + C = 0$ is $∣ A x_{0} + B y_{0} + C ∣/ A^{2} + B^{2}$ .

Proof. The vector $n = (A, B)$ is normal to the line. Parametrise the perpendicular through $(x_{0}, y_{0})$ as $(x, y) = (x_{0} + A t, y_{0} + B t)$ . Substituting into $A x + B y + C = 0$ gives

A (x_{0} + A t) + B (y_{0} + B t) + C = 0 ⟹ (A^{2} + B^{2}) t = - (A x_{0} + B y_{0} + C),

so $t = - (A x_{0} + B y_{0} + C) / (A^{2} + B^{2})$ . The distance from $(x_{0}, y_{0})$ to the foot of the perpendicular is $∣ t ∣ \cdot ∥ (A, B) ∥$ , hence

d = \frac{∣ A x _{0} + B y _{0} + C ∣}{A ^{2} + B ^{2}} \cdot A^{2} + B^{2} = \frac{∣ A x _{0} + B y _{0} + C ∣}{A ^{2} + B ^{2}} . □

Connections Master

Cartesian coordinates and distance 00.09.01. This unit deepens the coordinate-and-distance foundation of 00.09.01 by adding the full line-equation family (four interchangeable forms with conversions), the perpendicular-slope relation, and the circle as a locus. The distance formula proved there becomes the load-bearing tool here: the circle equation is the distance formula squared, and the midpoint becomes the center of a diameter-defined circle.
Linear equations and the line 00.03.01. The general line $A x + B y + C = 0$ studied abstractly in 00.03.01 acquires geometric content here through slope, intercepts, and the parallel-perpendicular classification. Solving a $2 \times 2$ linear system — the core skill of that unit — is exactly the tool used to find where a line meets a circle or where two perpendicular bisectors cross.
Quadratic equations and completing the square 00.03.02. Recovering the center and radius from $x^{2} + y^{2} + D x + E y + F = 0$ runs the completing-the-square technique of 00.03.02 twice, once per variable. The sign of the constant that remains decides whether the locus is a circle, a point, or empty, paralleling how the discriminant there decides the number of real roots.
Conic sections 00.10.01. The circle is the special case of an ellipse with coincident foci, and the focus-directrix definition of conics in 00.10.01 uses the same distance formula developed here. Completing the square in two variables, the central technique of this unit, is the entry point to classifying the general second-degree curve as parabola, ellipse, or hyperbola.

Historical & philosophical context Master

René Descartes published La Géométrie in 1637 as an appendix to the Discours de la méthode ^{[Descartes 1637]}, introducing the systematic correspondence between algebraic equations and geometric curves. By assigning numerical coordinates to points, Descartes showed that a curve could be studied through its equation: the distance and locus problems of classical Euclidean geometry became algebraic manipulations. Pierre de Fermat, working independently in the 1636 manuscript Ad Locos Planos et Solidos Isagoge (published posthumously in 1679) ^{[Fermat 1636]}, arrived at the same correspondence from the opposite direction, starting from geometric locus conditions and writing the equations they imply.

The two inventions differ in emphasis and complement each other. Descartes proceeds from equations to curves, using algebraic operations (including what we now call completing the square) to construct and classify curves. Fermat proceeds from geometric conditions — fixed distance from a point (a circle), fixed ratio of distances (a conic) — to the equations encoding them, which is precisely the locus viewpoint that defines the circle in this unit. The modern formulation of $R^{2}$ as a metric space, in which the distance formula is the canonical example of a metric and the circle is a metric sphere, crystallised with Fréchet's 1906 Sur quelques points du calcul fonctionnel and the twentieth-century development of normed vector spaces.

Bibliography Master

@book{Descartes1637,
  author = {Descartes, Ren{\'e}},
  title = {La G{\'e}om{\'e}trie},
  note = {Appendix to {\em Discours de la m{\'e}thode}, Leiden. Modern edition: Smith, D. E. and Latham, M. L. (transl.), Dover 1954},
  year = {1637}
}

@article{Fermat1636,
  author = {Fermat, Pierre de},
  title = {Ad Locos Planos et Solidos Isagoge},
  journal = {Varia Opera Mathematica},
  note = {Written 1636; published posthumously, Toulouse, 1679},
  year = {1636}
}

@book{Stewart2016,
  author = {Stewart, James and Redlin, Lothar and Watson, Saleem},
  title = {Precalculus: Mathematics for Calculus},
  edition = {7},
  publisher = {Cengage Learning},
  year = {2016}
}

@book{Sullivan2019,
  author = {Sullivan, Michael},
  title = {Precalculus},
  edition = {11},
  publisher = {Pearson},
  year = {2019}
}

@book{Apostol1967,
  author = {Apostol, Tom M.},
  title = {Calculus, Volume 1},
  publisher = {Wiley},
  year = {1967}
}

@article{Frechet1906,
  author = {Fr{\'e}chet, Maurice},
  title = {Sur quelques points du calcul fonctionnel},
  journal = {Rendiconti del Circolo Matematico di Palermo},
  volume = {22},
  pages = {1--74},
  year = {1906}
}

Prerequisites

00.09.01
00.03.01

Tier anchors

beginner: Stewart 2016 Precalculus (Cengage) Ch. 1 — the coordinate plane turns geometry into algebra: every point is a pair of numbers, every line and circle an equation.
intermediate: Stewart 2016 Precalculus (Cengage) Ch. 1–2, coordinate geometry and equations of circles; Sullivan Precalculus Ch. 2.
master: Descartes 1637 La Géométrie (Leiden) Book II; Fermat 1636 Ad Locos Planos et Solidos Isagoge.

References

Stewart — Precalculus: Mathematics for Calculus · Cengage (2016). Ch. 1: coordinates, the distance and midpoint formulas, line equations (slope-intercept, point-slope, two-point, general), parallel and perpendicular conditions.
Descartes — La Géométrie · 1637, Book II. The algebraic correspondence between curves and equations; the locus description of conics from coordinate conditions.
Fermat — Ad Locos Planos et Solidos Isagoge · 1636 manuscript, published 1679. Independent invention of coordinate geometry via the locus method.
Apostol — Calculus Vol. 1 · Ch. 1, coordinate systems, distance, and the analytic description of lines and circles.

Estimated time

beginner: 15m
intermediate: 40m
master: 80m