Sections of a Cone and the Circle

One Cone, Four Curves

Take a fixed vertical line $l$ (the axis) and a line $m$ meeting it at a point $V$ (the vertex) at a fixed angle $\alpha$. Rotating $m$ around $l$ sweeps out a double-napped right circular cone; each position of the rotating line is a generator, and the vertex separates the two nappes.

A conic section is the curve where a plane cuts this cone. Everything depends on the angle $\beta$ that the cutting plane makes with the axis:

Key Point (The angle dictionary): cutting away from the vertex, $\beta = 90°$ gives a circle; $\alpha < \beta < 90°$ gives an ellipse; $\beta = \alpha$ gives a parabola; $0 \le \beta < \alpha$ makes the plane cut both nappes — a hyperbola.

When the plane passes through the vertex, the sections degenerate: a point (when $\alpha < \beta \le 90°$), a straight line (when $\beta = \alpha$ — the degenerate parabola), or a pair of intersecting lines (when $0 \le \beta < \alpha$ — the degenerate hyperbola).

These four curves run planetary orbits, telescope mirrors, headlight reflectors and suspension-bridge cables — and the rest of the chapter gives each one a clean coordinate equation.

The Circle as a Locus

Key Point (Definition and equation): A circle is the set of all points at a fixed distance $r$ (the radius) from a fixed point $C(h, k)$ (the centre). Writing $CP = r$ with the distance formula and squaring:

$$(x - h)^2 + (y - k)^2 = r^2$$

Centre at the origin gives the familiar $x^2 + y^2 = r^2$. Centre $(-3, 2)$, radius 4 gives $(x + 3)^2 + (y - 2)^2 = 16$.

Reading centre and radius from an expanded equation. An equation like $x^2 + y^2 + 8x + 10y - 8 = 0$ hides its centre. Complete the square in $x$ and $y$ separately:

$$(x^2 + 8x + 16) + (y^2 + 10y + 25) = 8 + 16 + 25 \;\Rightarrow\; (x + 4)^2 + (y + 5)^2 = 49$$

so the centre is $(-4, -5)$ and the radius is 7.

Finding a circle from conditions. Two points on the circle plus a line containing the centre determine everything: the centre is equidistant from the two points (or lies on the perpendicular bisector of their chord), giving equations to solve simultaneously — the chord-bisector method worked through below.

[Board Tip] Signs are the classic trap: $(x + 4)^2$ means $h = -4$, so the centre is $(-4, -5)$, never $(4, 5)$. Say the standard form out loud as "$x$ minus $h$" before reading off coordinates.

Solved Examples

Example 1: Writing the equation directly

Find the equation of the circle with (i) centre $(0, 0)$ and radius $r$, (ii) centre $(-3, 2)$ and radius 4.

Solution:

Step 1 — (i) Centre at the origin. $h = k = 0$: $x^2 + y^2 = r^2$.

Step 2 — (ii) Substitute with signs. $h = -3$, $k = 2$: $(x - (-3))^2 + (y - 2)^2 = 4^2$, i.e. $(x + 3)^2 + (y - 2)^2 = 16$.

Takeaway: $h = -3$ turns the bracket into $(x + 3)$ — say "$x$ minus $h$" before substituting.

Example 2: A drill in both directions

Write the equations: (i) centre $(0, 2)$, radius 2; (ii) centre $\left(\frac{1}{2}, \frac{1}{4}\right)$, radius $\frac{1}{12}$; (iii) centre $(1, 1)$, radius $\sqrt{2}$; (iv) centre $(-a, -b)$, radius $\sqrt{a^2 - b^2}$.

Solution:

Step 1 — (i). $x^2 + (y - 2)^2 = 4$, which expands to $x^2 + y^2 - 4y = 0$ — through the origin.

Step 2 — (ii). $\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{4}\right)^2 = \frac{1}{144}$; multiplying by 144: $36x^2 + 36y^2 - 36x - 18y + 11 = 0$.

Step 3 — (iii). $(x - 1)^2 + (y - 1)^2 = 2$, i.e. $x^2 + y^2 - 2x - 2y = 0$ — also through the origin.

Step 4 — (iv). $(x + a)^2 + (y + b)^2 = a^2 - b^2$, i.e. $x^2 + y^2 + 2ax + 2by + 2b^2 = 0$.

Takeaway: Expanded forms whose constant is 0 pass through the origin — a quick structural read.

Example 3: Completing the square

Find the centre and radius of $x^2 + y^2 + 8x + 10y - 8 = 0$.

Solution:

Step 1 — Group and complete each square. $(x^2 + 8x + 16) + (y^2 + 10y + 25) = 8 + 16 + 25$.

Step 2 — Write the standard form. $(x + 4)^2 + (y + 5)^2 = 49 = 7^2$.

Step 3 — Read off. Centre $(-4, -5)$, radius 7.

Takeaway: Add HALF the coefficient, squared, to both sides — and remember $(x+4)^2$ means $h = -4$.

Example 4: Reading centre and radius

Find the centre and radius of: (i) $(x + 5)^2 + (y - 3)^2 = 36$; (ii) $x^2 + y^2 - 4x - 8y - 45 = 0$; (iii) $x^2 + y^2 - 8x + 10y - 12 = 0$; (iv) $2x^2 + 2y^2 - x = 0$.

Solution:

Step 1 — (i) Direct read. Centre $(-5, 3)$, radius 6.

Step 2 — (ii) Complete squares. $(x - 2)^2 + (y - 4)^2 = 45 + 4 + 16 = 65$: centre $(2, 4)$, radius $\sqrt{65}$.

Step 3 — (iii) Again. $(x - 4)^2 + (y + 5)^2 = 12 + 16 + 25 = 53$: centre $(4, -5)$, radius $\sqrt{53}$.

Step 4 — (iv) Normalise FIRST. Divide by 2: $x^2 + y^2 - \frac{x}{2} = 0$, so $\left(x - \frac{1}{4}\right)^2 + y^2 = \frac{1}{16}$: centre $\left(\frac{1}{4}, 0\right)$, radius $\frac{1}{4}$.

Takeaway: If $x^2$ and $y^2$ carry a coefficient, divide it out before completing any square.

Example 5: Two points and a centre line

Find the circle through $(2, -2)$ and $(3, 4)$ whose centre lies on $x + y = 2$.

Solution:

Step 1 — Equidistance condition. $(2 - h)^2 + (-2 - k)^2 = (3 - h)^2 + (4 - k)^2$.

Step 2 — Expand and simplify. The $h^2, k^2$ terms cancel: $8 - 4h + 4k = 25 - 6h - 8k$, i.e. $2h + 12k = 17$.

Step 3 — Solve with the centre line. Substitute $k = 2 - h$: $2h + 24 - 12h = 17$ gives $10h = 7$: $h = 0.7$, $k = 1.3$.

Step 4 — Radius. $r^2 = (2 - 0.7)^2 + (-2 - 1.3)^2 = 1.69 + 10.89 = 12.58$: the circle is $(x - 0.7)^2 + (y - 1.3)^2 = 12.58$.

Takeaway: "Equidistant from two points" plus "on a line" is always two linear equations in $h, k$.

Example 6: Chord bisector method

Find the circle through $(4, 1)$ and $(6, 5)$ whose centre lies on $4x + y = 16$.

Solution:

Step 1 — Perpendicular bisector of the chord. Midpoint $(5, 3)$; chord slope $\frac{5-1}{6-4} = 2$; bisector: $y - 3 = -\frac{1}{2}(x - 5)$, i.e. $x + 2y = 11$.

Step 2 — Intersect with the centre line. Solve $x + 2y = 11$ and $4x + y = 16$: multiply the second by 2 and subtract: $7x = 21$, so centre $(3, 4)$.

Step 3 — Radius and equation. $r^2 = (4-3)^2 + (1-4)^2 = 10$: $(x - 3)^2 + (y - 4)^2 = 10$, i.e. $x^2 + y^2 - 6x - 8y + 15 = 0$.

Takeaway: The centre always sits on the perpendicular bisector of every chord — one bisector plus the given line pins it.

Example 7: Same method, messier numbers

Find the circle through $(2, 3)$ and $(-1, 1)$ whose centre lies on $x - 3y - 11 = 0$.

Solution:

Step 1 — Bisector of the chord. Midpoint $\left(\frac{1}{2}, 2\right)$; chord slope $\frac{3-1}{2+1} = \frac{2}{3}$; bisector: $y - 2 = -\frac{3}{2}\left(x - \frac{1}{2}\right)$, i.e. $6x + 4y = 11$.

Step 2 — Solve with the centre line. $x - 3y = 11$ gives $x = 11 + 3y$; substituting: $66 + 18y + 4y = 11$, so $y = -\frac{5}{2}$, $x = \frac{7}{2}$.

Step 3 — Radius and equation. $r^2 = \left(2 - \frac{7}{2}\right)^2 + \left(3 + \frac{5}{2}\right)^2 = \frac{9}{4} + \frac{121}{4} = \frac{65}{2}$: expanding, $x^2 + y^2 - 7x + 5y - 14 = 0$.

Takeaway: Keep everything in fractions — the final expanded form clears them automatically.

Example 8: Radius given, centre constrained

Find the circle of radius 5 whose centre lies on the $x$-axis and which passes through $(2, 3)$.

Solution:

Step 1 — Name the centre. On the $x$-axis: $(h, 0)$.

Step 2 — Distance condition. $(2 - h)^2 + 3^2 = 25$ gives $(2 - h)^2 = 16$.

Step 3 — Both roots. $2 - h = \pm 4$: $h = -2$ or $h = 6$ — circles $(x + 2)^2 + y^2 = 25$ and $(x - 6)^2 + y^2 = 25$.

Takeaway: A squared condition means two centres — report both circles unless something excludes one.

Example 9: Through the origin; from centre and a point

(i) Find the circle through $(0, 0)$ making intercepts $a$ and $b$ on the axes. (ii) Find the circle with centre $(2, 2)$ passing through $(4, 5)$.

Solution:

Step 1 — (i) Use the general form. $x^2 + y^2 + Dx + Ey + F = 0$ through $(0,0)$ gives $F = 0$; through $(a, 0)$: $a^2 + Da = 0$, so $D = -a$; through $(0, b)$: $E = -b$.

Step 2 — (i) Write it. $x^2 + y^2 - ax - by = 0$.

Step 3 — (ii) Radius from the point. $r^2 = (4-2)^2 + (5-2)^2 = 13$: $(x - 2)^2 + (y - 2)^2 = 13$.

Takeaway: Three points on a circle → three linear equations in $D, E, F$ — the general form is built for this.

Example 10: Inside, on, or outside?

Does $(-2.5, 3.5)$ lie inside, outside or on the circle $x^2 + y^2 = 25$?

Solution:

Step 1 — Squared distance from the centre. $(-2.5)^2 + (3.5)^2 = 6.25 + 12.25 = 18.5$.

Step 2 — Compare with $r^2$. $18.5 < 25$.

Step 3 — Conclude. The point lies inside the circle.

Takeaway: Compare squared distances — no square roots needed for an inside/on/outside verdict.