DWG NO. 8.1 — Lesson 1 of 4

Circles and Ellipses

Unit 8: Conic Sections · ~25 min

Objective Write the standard-form equation of a circle or ellipse from its key features, and extract those features — center, radii, vertices, and foci — from a standard-form equation.

A circle is the simplest conic: every point the same distance from a center. An ellipse relaxes that rule slightly — instead of one fixed point, it uses two, called foci (singular: focus), and instead of a fixed distance, it fixes the sum of the two distances to those foci. Squash the two foci together into a single point and the sum-of-distances rule collapses back into the circle's single-distance rule — the circle is really just an ellipse whose foci coincide.

Circles

A circle is the set of all points at a fixed distance r (the radius) from a center (h, k).

Standard form (x − h)² + (y − k)² = r², center (h, k), radius r

A circle's general form, x² + y² + Dx + Ey + F = 0, hides the center and radius inside the coefficients. Completing the square on both x and y recovers standard form — the same move you used on quadratics back in Unit 2, just applied twice.

Ellipses

An ellipse is the set of all points P for which the sum of the distances from P to two fixed foci is constant. That constant sum turns out to equal 2a, where a is the distance from the center to a vertex along the ellipse's longer axis (the major axis). The shorter axis is the minor axis, with half-length b.

OrientationStandard formFoci & vertices lie on
Major axis horizontal(x−h)\(\frac{^{2}}{a^{2}}\) + (y−k)\(\frac{^{2}}{b^{2}}\) = 1the horizontal line y = k
Major axis vertical(x−h)\(\frac{^{2}}{b^{2}}\) + (y−k)\(\frac{^{2}}{a^{2}}\) = 1the vertical line x = h

In both cases a > b, and the larger denominator tells you which axis is major. The distance from center to focus, c, satisfies c² = a² − b² — the foci always sit strictly inside the vertices, closer to the center.

F₁ F₂ (−5,0) (5,0) (0,3) (0,−3)

Ellipse \(\frac{x^{2}}{25}\) + \(\frac{y^{2}}{9}\) = 1 — a = 5, b = 3, c = \(\sqrt{25-9}\) = 4, foci at (±4, 0)

Worked Example 1 · Circle from general form
ProblemFind the center and radius of x² + y² − 6x + 4y − 3 = 0.
1Group and complete the square: (x² − 6x) + (y² + 4y) = 3.
2(x² − 6x + 9) + (y² + 4y + 4) = 3 + 9 + 4.
3(x − 3)² + (y + 2)² = 16.
Center (3, −2), radius 4
Worked Example 2 · Features of an ellipse
ProblemFind the vertices, co-vertices, and foci of (x+1)\(\frac{^{2}}{25}\) + (y−2)\(\frac{^{2}}{9}\) = 1.
1Center: (−1, 2). The larger denominator (25) is under x, so the major axis is horizontal: a²=25 → a=5, b²=9 → b=3.
2Vertices: (−1±5, 2) = (4, 2) and (−6, 2).
3Co-vertices: (−1, 2±3) = (−1, 5) and (−1, −1).
4c² = a² − b² = 25 − 9 = 16 → c = 4. Foci: (−1±4, 2) = (3, 2) and (−5, 2).
Vertices (4,2), (−6,2) · Foci (3,2), (−5,2)

Guided practice