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Lesson 1 of 5

Conic Sections · Lesson 1 of 5

Conic Sections and Circles

“See how cutting a cone produces curves, then build and interpret circle equations from distance.”

Learning Objectives

• Identify the axis, vertex, generators and nappes of a double cone, and classify its plane sections. • Distinguish ordinary conics from point, line and intersecting-line degeneracies. • Derive a circle equation from its centre and radius, and recover those features by completing squares. • Use point and centre conditions to construct a circle and check whether a point lies inside, on or outside it.

From straight lines to curved paths

A straight line is described by a linear equation. A curved path can also be described by an equation: the equation tells us exactly which points belong to the path. In this chapter we meet four related curves—circle, parabola, ellipse and hyperbola—and discover that each equation comes from a simple distance condition.

The word locus means the set of all points satisfying a stated condition. For example, all points exactly 4 units from a fixed point form a circle. This viewpoint is more useful than memorising a sketch: if we translate the condition into coordinates, we can reason about an entire curve at once. These curves appear in reflector design, telescope and antenna shapes, and models of planetary paths.

A double cone and its plane sections

Imagine a fixed vertical line and a second line meeting it at a fixed angle. Rotate the second line around the vertical line while keeping that angle unchanged. The rotating line sweeps out a cone extending indefinitely in both directions. The familiar cone of an ice-cream cone is only one portion of this double cone.

Definition
Parts of a double-napped right circular cone

The fixed meeting point is the vertex. The fixed vertical line is the axis. A rotating slant line is a generator. The two portions separated by the vertex are the nappes. A conic section is the intersection of this conical surface with a plane.

Let α be the acute angle between a generator and the axis. Let β be the angle between the cutting plane and the axis; a plane perpendicular to the axis has β = 90°. First keep the plane away from the vertex. The plane’s tilt decides whether its intersection is a closed curve, an open curve on one nappe, or two branches on both nappes.

Position of a plane avoiding the vertexAngle conditionIntersection
Perpendicular to the axisβ = 90°Circle
Tilted, cutting entirely across one nappeα < β < 90°Ellipse
Parallel to a generatorβ = αParabola
Cuts both nappes0 ≤ β < αHyperbola
CircleEllipseParabolaHyperbolaAxial schematics: blue lines show the cutting planes’ traces.
How plane tilt changes the cone section— These are cross-sectional schematics, not complete 3D intersections. A horizontal, tilted, generator-parallel or axis-parallel plane produces the four profiles shown next.
CircleEllipseParabolaHyperbolaOne nappe: circle, ellipse or parabola. Both nappes: hyperbola.
Four conic profiles— These are plane views of the resulting intersections; the hyperbola has two branches.
Vertex VAxisGeneratorUpper nappeLower nappe
An indefinitely extending double cone— The sketch shows only a finite portion; each nappe continues beyond its drawn rim.

Now move the cutting plane through the vertex. The intersection can collapse: for α < β ≤ 90° it is only the vertex; for β = α it is one generator line; for 0 ≤ β < α it is a pair of intersecting generator lines. These are degenerate conic sections. A degenerate section is still a meaningful intersection, but it is not one of the ordinary curved shapes.

PointOne lineIntersecting linesThrough-vertex sections collapse according to the plane’s tilt.
Degenerate conic sections— When the plane passes through the vertex the intersection can be a point, one line or two intersecting lines.
Example — Classifying a cut

Problem
A cone has α = 30°. Classify planes avoiding the vertex with β = 90°, 50°, 30° and 10°. What changes if they pass through the vertex?

  1. 1.At 90° the plane is perpendicular to the axis, so the section is a circle. At 50°, α < β < 90°, so it is an ellipse.
  2. 2.At 30° the plane is parallel to a generator, producing a parabola. At 10°, β < α, so it cuts both nappes and produces a hyperbola.
  3. 3.Through the vertex, the first two cuts give a point, the 30° cut gives one straight line, and the 10° cut gives two intersecting lines. The position of the plane matters as well as its tilt.

A circle is a fixed-distance locus

Choose a point C(h, k) and a positive length r. A point P(x, y) belongs to the circle precisely when CP = r. The numbers h and k locate the centre; they do not measure the size of the circle. The radius r measures its size.

Definition
Circle, centre and radius

A circle is the set of all points in a plane at the same positive distance from a fixed point. The fixed point is the centre and the distance is the radius.

The horizontal difference between C and P is x − h, and the vertical difference is y − k. These form the legs of a right triangle whose hypotenuse is CP. The distance formula gives CP = √((x − h)² + (y − k)²). Equating this nonnegative distance to r and squaring gives the standard circle equation.

Circle with centre (h, k) and radius rLaTeX
Here r > 0. Every point on the circle satisfies this equation, and every point satisfying it has distance r from the centre.

The converse matters: if the equation holds, taking the nonnegative square root gives CP = r, so no additional points have slipped in during squaring. If h = k = 0, the equation simplifies to x² + y² = r². This is the origin-centred circle, not a separate rule to memorise.

C(h, k)P(x, y)rx − hy − kxy
Distance builds the circle equation— The two coordinate differences are perpendicular legs; the radius is their hypotenuse.
Example — An origin-centred circle

Problem
Find the equation of a circle centred at (0, 0) with radius 6.

  1. 1.Use h = 0, k = 0 and r = 6 in the distance condition.
  2. 2.The equation is (x − 0)² + (y − 0)² = 6², so x² + y² = 36.
  3. 3.For a quick check, (6, 0) satisfies it and is 6 units from the origin; (0, 0) does not satisfy it and is the centre, not a point on the circle.
Example — A centre with a negative coordinate

Problem
Find the equation of a circle with centre (−3, 2) and radius 4.

  1. 1.Use x − h = x − (−3) = x + 3 and y − k = y − 2.
  2. 2.Square the radius: r² = 16. Therefore (x + 3)² + (y − 2)² = 16.
  3. 3.The signs inside the brackets are opposite to the signs of the centre coordinates. At the centre the left side is zero, whereas on the circle it is 16.

Recovering a centre by completing squares

A circle equation may arrive expanded. Completing a square reverses the expansion and reveals the centre and radius. Whatever quantity we add on the left, we must also add on the right; otherwise we change the curve.

For example, x² + 8x becomes (x + 4)² only after adding 16, because (x + 4)² = x² + 8x + 16. Similarly, y² + 10y needs 25. If the coefficients of x² and y² are the same positive number, divide by that number first. The standard form has coefficient 1 on each square.

Example — Completing both squares

Problem
Find the centre and radius of x² + y² + 8x + 10y − 8 = 0.

  1. 1.Move the constant: (x² + 8x) + (y² + 10y) = 8.
  2. 2.Half of 8 is 4, whose square is 16. Half of 10 is 5, whose square is 25. Add 16 + 25 to both sides.
  3. 3.(x + 4)² + (y + 5)² = 8 + 16 + 25 = 49.
  4. 4.The centre is (−4, −5) and the positive radius is √49 = 7, not 49.
Example — Normalise before completing squares

Problem
Find the centre and radius of 2x² + 2y² − x = 0.

  1. 1.Divide by 2: x² + y² − x/2 = 0.
  2. 2.Complete x² − x/2 by adding 1/16 to both sides: (x − 1/4)² + y² = 1/16.
  3. 3.The centre is (1/4, 0) and the radius is 1/4. This small positive radius follows from the square root of the right side.

Completing squares can also expose an impossible claim. A sum of real squares cannot equal a negative number. A right side of zero describes a single point rather than an ordinary circle. Thus a real circle needs a strictly positive value of r².

The centre and radius are not read directly from coefficients

In (x + 3)² + (y − 2)² = 16, the centre is (−3, 2) and r = 4. Expanded linear coefficients do not directly give the centre: first complete squares or derive the equivalent relation.

Constructing a circle from conditions

Sometimes the centre and radius are not supplied directly. Write a centre as (h, k), translate each given condition into an equation, then solve. If two points lie on the same circle, their squared distances from its centre are equal. Subtracting these equations cancels h², k² and r², leaving a linear relation.

Example — Two points and a centre line

Problem
Find a circle through (2, −2) and (3, 4) whose centre lies on x + y = 2.

  1. 1.Write the centre as (h, k). The point conditions are (2 − h)² + (−2 − k)² = r² and (3 − h)² + (4 − k)² = r². The centre-line condition is h + k = 2.
  2. 2.Expand and subtract the first point equation from the second: 17 − 2h − 12k = 0, so 2h + 12k = 17.
  3. 3.Put h = 2 − k: 4 − 2k + 12k = 17. Hence k = 13/10 and h = 7/10.
  4. 4.Using (2, −2), r² = (13/10)² + (−33/10)² = 1258/100 = 629/50.
  5. 5.Therefore (x − 7/10)² + (y − 13/10)² = 629/50. Both supplied points satisfy it, and 7/10 + 13/10 = 2 confirms the centre constraint.
Example — A condition with two valid circles

Problem
A circle has radius 5, its centre is on the x-axis, and it passes through (2, 3). Find all possible equations.

  1. 1.Write the centre as (h, 0). The point condition is (2 − h)² + 3² = 25.
  2. 2.(2 − h)² = 16, so 2 − h = 4 or −4. Thus h = −2 or 6.
  3. 3.The equations are (x + 2)² + y² = 25 and (x − 6)² + y² = 25. Retain both signs: the information does not select just one centre.
Example — A circle through the origin with axis intercepts

Problem
A circle passes through (0, 0), (u, 0) and (0, v), where u and v are nonzero signed intercepts. Find its equation.

  1. 1.Expand the standard form into x² + y² + Dx + Ey + F = 0. The origin gives F = 0.
  2. 2.At (u, 0), u² + Du = 0, giving D = −u. At (0, v), v² + Ev = 0, giving E = −v.
  3. 3.The equation is x² + y² − ux − vy = 0.
  4. 4.Completing squares gives (x − u/2)² + (y − v/2)² = (u² + v²)/4. The centre is (u/2, v/2) and the radius is √(u² + v²)/2.

Inside, on or outside

A circle encloses points whose distances from the centre are less than the radius. We can compare squared distances because both distance and radius are nonnegative. This avoids an unnecessary square root and connects point classification directly to the circle equation.

Squared distance d² = (x₁ − h)² + (y₁ − k)²Position
d² < r²Inside
d² = r²On
d² > r²Outside
Example — Checking a point

Problem
Does (−2.5, 3.5) lie inside, on or outside x² + y² = 25?

  1. 1.The centre is the origin and r² = 25.
  2. 2.The squared distance is (−2.5)² + 3.5² = 6.25 + 12.25 = 18.5.
  3. 3.Since 18.5 < 25, the point lies inside. It need not satisfy the circle equation: only boundary points do.

Check your understanding

Use these questions to connect the cone picture with the distance equation. Pay attention to the plane’s position, signs inside squared brackets, and the difference between a radius and a squared radius.

Quiz

Quick check

A plane avoiding the vertex is parallel to a generator. Which section results?

Quick check

A plane through the vertex has 0 ≤ β < α. What is its intersection?

Quick check

What are the centre and radius of (x − 2)² + (y + 3)² = 25?

Quick check

Which equation has centre (−1, 4) and radius 3?

Quick check

What is the radius of x² + y² − 6x + 2y − 6 = 0?

Quick check

Where is (3, 4) relative to x² + y² = 25?

Practice Problems

Practice Problems
  1. Explain why a plane cutting both nappes produces a hyperbola, while one perpendicular to the axis produces a circle. Include the assumption that the plane avoids the vertex.
  2. Derive the equation of a circle with centre (−2, 3) and radius 4 from the distance formula.
  3. Recover the centre and radius of x² + y² − 4x − 8y − 45 = 0, showing each square-completion step.
  4. Find the circle through (4, 1) and (6, 5) whose centre lies on 4x + y = 16.
  5. Find every circle of radius 5 with centre on the x-axis passing through (2, 3), and explain why there are two.
  6. Find the circle centred at (2, 2) through (4, 5), then classify (2, 4) relative to it.
  7. For centre (−p, −q) and proposed radius √(p² − q²), state the parameter condition for an ordinary real circle and write its equation.

Key Takeaways

Key Takeaways

• Cone sections depend on both the tilt of the plane and whether it passes through the vertex. • A circle is a fixed positive-distance locus: (x − h)² + (y − k)² = r². • Completing squares reveals the centre and positive radius; equal quadratic coefficients should first be normalised. • Translate each point or centre condition into an equation and retain every valid solution. • Compare squared distances with r² to distinguish inside, boundary and outside points.

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Next · Lesson 2

Parabolas: Focus, Directrix, and Orientation