Chapter 10 · Conic Sections

10.1 Sections of a Cone and the Circle

A plane cutting a double cone makes a circle, an ellipse, a parabola or a hyperbola — the angle of the cut decides which. The simplest of them, the circle with centre $(h,k)$ and radius $r$, has equation $(x-h)^2+(y-k)^2=r^2$.

Switch on a torch in a dark room and point it straight at a wall: the patch of light is a circle. Tilt the torch a little and the circle stretches into an oval — an ellipse. Tilt it further until the top edge of the beam runs parallel to the wall and the patch no longer closes: its edge is a parabola. Tilt further still and the edge becomes one branch of a hyperbola. The beam of a torch is a cone of light, and the wall is a flat plane cutting it; the four curves you see are the conic sections.

In this section we make that experiment precise — which angle of cut gives which curve — and then study the first and most familiar conic, the circle, with coordinates. A circle is fixed by its centre and radius, and the distance formula turns that description into an equation. Going the other way, we will learn to look at an equation such as $x^2+y^2-6x+10y-2=0$ and, by completing the square, read off its centre and radius. The same "complete the square and read off the shape" habit will serve you for the parabola, ellipse and hyperbola in the rest of the chapter.

Slicing a Double Cone

📘 Definition 10.1 — Double cone

Let $l$ be a fixed vertical line and let $m$ be another line that meets $l$ at a fixed point $V$ and makes a fixed acute angle $\alpha$ with it. Rotate $m$ about $l$, keeping the angle $\alpha$ fixed. The surface swept out by $m$ is a double cone (a double-napped right circular cone). The point $V$ is its vertex, the line $l$ is its axis, each position of the rotating line $m$ is a generator of the cone, and $\alpha$ is the semi-vertical angle. The two pieces of the surface above and below $V$ are the two nappes.

A cone in everyday life (an ice-cream cone, a traffic cone) is one nappe cut off at some height. The mathematical double cone goes on for ever in both directions. Now cut it with a plane and let $\beta$ be the angle between the plane and the axis $l$, with $0^\circ\le\beta\le90^\circ$. Comparing $\beta$ with $\alpha$ decides the shape of the cut.

📘 Definition 10.2 — Conic sections

The curve in which a plane cuts a double cone is a conic section (or simply a conic). If the plane does not pass through the vertex $V$:

$\beta=90^\circ$ (plane perpendicular to the axis): a circle;
$\alpha\lt\beta\lt90^\circ$: an ellipse — the plane cuts right across one nappe;
$\beta=\alpha$ (plane parallel to a generator): a parabola — the curve does not close up;
$0^\circ\le\beta\lt\alpha$: a hyperbola — the plane cuts both nappes, so the curve has two separate branches.

If the plane does pass through $V$, we get a degenerate conic: a single point (when $\alpha\lt\beta\le90^\circ$), a single line, which is a generator (when $\beta=\alpha$), or a pair of intersecting lines (when $0^\circ\le\beta\lt\alpha$).

A useful way to remember the list: think of the plane being tilted steadily from horizontal ($\beta=90^\circ$) towards vertical ($\beta=0^\circ$). As long as the plane is steeper than flat but flatter than the cone's side it slices the nappe in a closed loop. At the instant it becomes parallel to the side ($\beta=\alpha$) the loop breaks open, and once it is steeper than the side it reaches down into the other nappe too. The degenerate cases are what the same three cuts become when the plane is slid along until it passes through the vertex. In this course the degenerate conics are only named; all the work is on the four curves.

✏️ Example 1 — Naming the section

A double cone has semi-vertical angle $\alpha=35^\circ$. Name the section made by a plane that makes an angle $\beta$ with the axis, where
(a) $\beta=90^\circ$; (b) $\beta=60^\circ$; (c) $\beta=35^\circ$; (d) $\beta=20^\circ$, the plane not passing through the vertex in (a)–(d);
(e) the plane passes through the vertex and $\beta=70^\circ$; (f) through the vertex with $\beta=10^\circ$; (g) through the vertex with $\beta=35^\circ$.

💡 Show solution▶

Rule Compare $\beta$ with $\alpha=35^\circ$: $\beta=90^\circ$ circle; $35^\circ\lt\beta\lt90^\circ$ ellipse; $\beta=35^\circ$ parabola; $\beta\lt35^\circ$ hyperbola.

(a)–(d) (a) circle; (b) $60^\circ\gt35^\circ$: ellipse; (c) $\beta=\alpha$: parabola; (d) $20^\circ\lt35^\circ$: hyperbola.

(e)–(g) Through the vertex: (e) $70^\circ\gt35^\circ$, a single point; (f) $10^\circ\lt35^\circ$, a pair of intersecting lines; (g) $\beta=\alpha$, a single line (the plane touches the cone along a generator).

Check In (d) and (f) the plane is steeper than the cone's side, so it meets both nappes — two branches, or two lines crossing at $V$ ✓.

Left: the double cone seen from the side, with the cutting plane edge-on (purple line) at angle $\beta$ to the axis. Right: the true shape of the cut, as you would see it looking straight at the plane. Press Play to tilt the plane from $\beta=90^\circ$ down to $0^\circ$ and back, and watch circle → ellipse → parabola → hyperbola. Move the plane's offset to $0$ to push it through the vertex and see the degenerate cases.

$\beta$60°$\alpha$30°offset1

$\beta$ compared with $\alpha$—

section—

nappes cut—

The Circle with Centre $(h,k)$

📘 Definition 10.3 — Circle

A circle is the set of all points in a plane that are at a fixed distance from a fixed point of the plane. The fixed point is the centre and the fixed distance is the radius $r$ ($r\gt0$).

📐 Theorem 10.1 — Equation of a circle

The circle with centre $C(h,k)$ and radius $r$ has equation $$(x-h)^2+(y-k)^2=r^2.$$ In particular, the circle with centre at the origin and radius $r$ is $x^2+y^2=r^2$.

Proof. A point $P(x,y)$ lies on the circle exactly when $CP=r$. By the distance formula this says $\sqrt{(x-h)^2+(y-k)^2}=r$. Both sides are non-negative, so squaring loses nothing: $CP=r$ holds if and only if $(x-h)^2+(y-k)^2=r^2$. Thus the points satisfying the equation are exactly the points of the circle. Putting $h=k=0$ gives the second form. $\blacksquare$

Notice the signs: the centre $(h,k)$ appears in the equation with minus signs. The circle $(x+3)^2+(y-2)^2=16$ is $(x-(-3))^2+(y-2)^2=4^2$, so its centre is $(-3,2)$ and its radius is $4$ (not $16$).

✏️ Example 2 — Writing down the equation

Find the equation of the circle with
(a) centre $(0,0)$ and radius $4$;
(b) centre $(-3,2)$ and radius $4$;
(c) centre $\left(\tfrac23,-\tfrac12\right)$ and radius $\tfrac56$;
(d) centre $(-3,4)$, touching the $x$-axis.

💡 Show solution▶

(a) $x^2+y^2=16$.

(b) $(x+3)^2+(y-2)^2=16$, which expands to $x^2+y^2+6x-4y-3=0$.

(c) $\left(x-\tfrac23\right)^2+\left(y+\tfrac12\right)^2=\tfrac{25}{36}$. Expanding, $x^2+y^2-\tfrac43x+y+\tfrac49+\tfrac14-\tfrac{25}{36}=0$; since $\tfrac{16+9-25}{36}=0$, this is $x^2+y^2-\tfrac43x+y=0$, i.e. $3x^2+3y^2-4x+3y=0$. (The constant term is $0$, so the circle passes through the origin.)

(d) The radius to the point of contact is perpendicular to the $x$-axis, so it equals the distance of the centre from the $x$-axis: $r=|4|=4$. The circle is $(x+3)^2+(y-4)^2=16$.

Check (c) The origin: $\left(\tfrac23\right)^2+\left(\tfrac12\right)^2=\tfrac{16+9}{36}=\tfrac{25}{36}=r^2$ ✓.

✏️ Example 3 — A circle on a given diameter

The points $A(-2,5)$ and $B(4,-3)$ are the ends of a diameter of a circle. Find its equation.

💡 Show solution▶

Centre The centre is the mid-point of $AB$: $\left(\tfrac{-2+4}{2},\tfrac{5-3}{2}\right)=(1,1)$.

Radius $r=CA=\sqrt{(-2-1)^2+(5-1)^2}=\sqrt{9+16}=5$ (half of $AB=\sqrt{36+64}=10$).

Answer $(x-1)^2+(y-1)^2=25$, i.e. $x^2+y^2-2x-2y-23=0$.

Check $B$: $(4-1)^2+(-3-1)^2=9+16=25$ ✓. A second route: a point $P$ is on the circle exactly when $\angle APB=90^\circ$, which gives $(x+2)(x-4)+(y-5)(y+3)=0$ — expand it and you get the same equation.

The General Equation and Completing the Square

Expand $(x-h)^2+(y-k)^2=r^2$ and every circle takes the shape $$x^2+y^2+2gx+2fy+c=0,$$ with $g=-h$, $f=-k$ and $c=h^2+k^2-r^2$. Two features mark it out: the coefficients of $x^2$ and $y^2$ are equal, and there is no $xy$ term. The question is the reverse one: given such an equation, does it describe a circle, and if so, which?

📐 Theorem 10.2 — The general equation of a circle

Let $g,f,c$ be real numbers. The equation $$x^2+y^2+2gx+2fy+c=0$$ can be written as $(x+g)^2+(y+f)^2=g^2+f^2-c$. Hence:

if $g^2+f^2-c\gt0$, it is a circle with centre $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$;
if $g^2+f^2-c=0$, the only point satisfying it is $(-g,-f)$ (a point circle);
if $g^2+f^2-c\lt0$, no point of the plane satisfies it.

Proof. Complete the square in $x$ and in $y$: $x^2+2gx=(x+g)^2-g^2$ and $y^2+2fy=(y+f)^2-f^2$. The equation becomes $(x+g)^2+(y+f)^2=g^2+f^2-c$. The left side is the square of the distance from $(x,y)$ to $(-g,-f)$. If the right side is positive, call it $r^2$; Theorem 10.1 gives the circle. If it is $0$, the distance must be $0$, so $(x,y)=(-g,-f)$. If it is negative, no distance squared can equal it. $\blacksquare$

🧭 Finding the centre and radius

1. Check that $x^2$ and $y^2$ have the same coefficient and that there is no $xy$ term; divide through by that coefficient so that both become $1$.
2. Group the $x$ terms and the $y$ terms, and move the constant to the right.
3. Complete each square by adding (half the coefficient)$^2$ to both sides.
4. Read off the centre (change the signs inside the brackets) and the radius (square root of the right side, if it is positive).

✏️ Example 4 — Centre and radius from an equation

Decide whether each equation is a circle; if it is, find its centre and radius.
(a) $x^2+y^2-6x+10y-2=0$ (b) $2x^2+2y^2-4x+12y+11=0$ (c) $x^2+y^2+4x-2y+6=0$ (d) $x^2+y^2+8x-6y+25=0$

💡 Show solution▶

(a) $(x^2-6x+9)+(y^2+10y+25)=2+9+25$, so $(x-3)^2+(y+5)^2=36$: centre $(3,-5)$, radius $6$.

(b) Divide by $2$: $x^2+y^2-2x+6y+\tfrac{11}{2}=0$. Then $(x-1)^2+(y+3)^2=1+9-\tfrac{11}{2}=\tfrac92$: centre $(1,-3)$, radius $\sqrt{9/2}=\dfrac{3}{\sqrt2}=\dfrac{3\sqrt2}{2}\approx2.12$.

(d) $(x+4)^2+(y-3)^2=16+9-25=0$: only the point $(-4,3)$ — a point circle, not a genuine circle.

Check (a) by Theorem 10.2: $g=-3$, $f=5$, $c=-2$, so the centre is $(-g,-f)=(3,-5)$ and $r=\sqrt{9+25+2}=6$ ✓.

✏️ Example 5 — The circle through three points

Find the equation of the circle through $A(5,3)$, $B(-2,2)$ and $C(6,-4)$, and its centre and radius.

💡 Show solution▶

Set up Take $x^2+y^2+2gx+2fy+c=0$ and substitute each point:
$A$: $34+10g+6f+c=0$; $B$: $8-4g+4f+c=0$; $C$: $52+12g-8f+c=0$.

Eliminate $c$ $A-B$: $26+14g+2f=0$, i.e. $7g+f=-13$. $C-A$: $18+2g-14f=0$, i.e. $g-7f=-9$.

Solve From the first, $f=-13-7g$; then $g+91+49g=-9$, so $g=-2$ and $f=1$. From $B$: $c=-8+4g-4f=-8-8-4=-20$.

Answer $x^2+y^2-4x+2y-20=0$: centre $(2,-1)$, radius $\sqrt{4+1+20}=5$.

Check Distances from $(2,-1)$: to $A$, $\sqrt{9+16}=5$; to $B$, $\sqrt{16+9}=5$; to $C$, $\sqrt{16+9}=5$ ✓.

✏️ Example 6 — Two points and a line for the centre

Find the equation of the circle that passes through $P(0,-2)$ and $Q(7,5)$ and whose centre lies on the line $x-y=1$.

💡 Show solution▶

Idea Let the centre be $(h,k)$. It is on the line, so $h-k=1$; and it is equally far from $P$ and $Q$.

Equal distances $h^2+(k+2)^2=(h-7)^2+(k-5)^2$. The $h^2$ and $k^2$ cancel: $4k+4=-14h+49-10k+25$, so $14h+14k=70$, i.e. $h+k=5$.

Solve $h-k=1$ and $h+k=5$ give $h=3$, $k=2$. Then $r^2=3^2+(2+2)^2=25$.

Answer $(x-3)^2+(y-2)^2=25$, i.e. $x^2+y^2-6x-4y-12=0$. Check $Q$: $16+9=25$ ✓.

⚠️ Common mistakes with circles

1. The wrong signs for the centre. $(x-3)^2+(y+5)^2=36$ has centre $(3,-5)$; in $x^2+y^2+2gx+2fy+c=0$ the centre is $(-g,-f)$ — half the coefficients of $x$ and $y$, with their signs changed.

2. Reading $r^2$ as $r$. $(x-1)^2+y^2=16$ has radius $4$, not $16$.

3. Forgetting to divide first. In $2x^2+2y^2-4x+12y+11=0$ the centre is $(1,-3)$, not $(2,-6)$: make the coefficients of $x^2$ and $y^2$ equal to $1$ before reading off $g$ and $f$.

4. Calling every quadratic a circle. $x^2+2y^2=4$ (unequal coefficients) and $x^2+xy+y^2=1$ (an $xy$ term) are not circles; and $x^2+y^2+4x-2y+6=0$ looks like one but has no points at all. Always check the sign of $g^2+f^2-c$.

5. Completing the square on one side only. Whatever you add to the left ($9$ and $25$ in Example 4(a)) must be added to the right as well.

🚀 Beyond the board syllabus (JEE Main · JEE Advanced)

In JEE Main 2026 (points of intersection of a line and a circle centred at the origin) and JEE Advanced 2026 (tangents; a circle meeting a straight line); not in the CBSE year-end exam. Tangency conditions are a favourite of both. When does the line $y=mx+c$ touch the circle $x^2+y^2=a^2$ (meet it at exactly one point)?

Method. Substitute $y=mx+c$: $x^2+(mx+c)^2=a^2$, i.e. $(1+m^2)x^2+2mcx+(c^2-a^2)=0$. The line touches the circle exactly when this quadratic has equal roots, i.e. its discriminant is zero: $4m^2c^2-4(1+m^2)(c^2-a^2)=0$. This simplifies to $a^2(1+m^2)-c^2=0$: $$y=mx+c \text{ touches } x^2+y^2=a^2\iff c^2=a^2(1+m^2).$$ (If $c^2\lt a^2(1+m^2)$ the line cuts the circle in two points; if $c^2\gt a^2(1+m^2)$ it misses it.)

Worked example. For which $c$ does $y=x+c$ touch $x^2+y^2=8$? Solution. $m=1$, $a^2=8$: $c^2=8\cdot2=16$, so $c=\pm4$. Check with $c=4$: $x^2+(x+4)^2=8$ gives $2x^2+8x+8=0$, i.e. $(x+2)^2=0$ — a single point $(-2,2)$ ✓.

Exercises

Write the equation of the circle, in the form $(x-h)^2+(y-k)^2=r^2$ and expanded.

1 Centre $(0,2)$, radius $2$.
2 Centre $(-2,3)$, radius $4$.
3 Centre $\left(-\tfrac12,\tfrac13\right)$, radius $\tfrac12$.
4 Centre $(p,-p)$, passing through the origin ($p\ne0$).
5 The ends of a diameter are $(3,-1)$ and $(-5,7)$.
6 Centre $(2,-3)$, passing through $(5,1)$.
7 Centre $(-4,-3)$, touching the $y$-axis.
8 Radius $3$, lying in the first quadrant and touching both axes.
9 Same centre as $x^2+y^2-4x+6y-3=0$ (concentric with it) and passing through the origin.

💡 Show solutions▶

1 $x^2+(y-2)^2=4$, i.e. $x^2+y^2-4y=0$.

2 $(x+2)^2+(y-3)^2=16$, i.e. $x^2+y^2+4x-6y-3=0$.

3 $\left(x+\tfrac12\right)^2+\left(y-\tfrac13\right)^2=\tfrac14$; expanding, $x^2+y^2+x-\tfrac23y+\tfrac19=0$, i.e. $9x^2+9y^2+9x-6y+1=0$.

4 $r^2=p^2+p^2=2p^2$: $(x-p)^2+(y+p)^2=2p^2$, i.e. $x^2+y^2-2px+2py=0$.

5 Centre $(-1,3)$, $r^2=(3+1)^2+(-1-3)^2=32$: $(x+1)^2+(y-3)^2=32$, i.e. $x^2+y^2+2x-6y-22=0$.

6 $r=\sqrt{9+16}=5$: $(x-2)^2+(y+3)^2=25$, i.e. $x^2+y^2-4x+6y-12=0$.

7 $r=$ distance from the $y$-axis $=4$: $(x+4)^2+(y+3)^2=16$, i.e. $x^2+y^2+8x+6y+9=0$.

8 Centre $(3,3)$: $(x-3)^2+(y-3)^2=9$, i.e. $x^2+y^2-6x-6y+9=0$.

9 Centre $(2,-3)$, $r^2=4+9=13$: $(x-2)^2+(y+3)^2=13$, i.e. $x^2+y^2-4x+6y=0$.

Centres, radii and the general equation.

1 Find the centre and radius of $(x+5)^2+(y-3)^2=36$.
2 Find the centre and radius of $x^2+y^2-4x-8y-45=0$.
3 Find the centre and radius of $x^2+y^2-8x+10y-12=0$.
4 Find the centre and radius of $3x^2+3y^2+12x-18y+3=0$.
5 Find the centre and radius of $4x^2+4y^2-4x+8y-11=0$.
6 Find the centre and radius of $x^2+y^2-10x=0$, and explain why it passes through the origin.
7 For which values of $k$ does $x^2+y^2+6x-4y+k=0$ represent a circle? For which $k$ is its radius $2$?
8 Describe the set of points satisfying (a) $x^2+y^2-2x+4y+5=0$; (b) $x^2+y^2+2x+2y+3=0$.
9 Does the point $(3,4)$ lie inside, on or outside the circle $x^2+y^2-4x-2y-11=0$?

Check: centre and radius of $x^2+y^2+Dx+Ey+F=0$

$D$$E$$F$

$g=\tfrac D2,\ f=\tfrac E2$—

centre—

$r^2$ and $r$—

💡 Show solutions▶

1 Centre $(-5,3)$, radius $6$.

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

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

4 Divide by $3$: $x^2+y^2+4x-6y+1=0$, so $(x+2)^2+(y-3)^2=4+9-1=12$: centre $(-2,3)$, radius $2\sqrt3$.

5 Divide by $4$: $x^2+y^2-x+2y-\tfrac{11}{4}=0$, so $\left(x-\tfrac12\right)^2+(y+1)^2=\tfrac14+1+\tfrac{11}4=4$: centre $\left(\tfrac12,-1\right)$, radius $2$.

6 $(x-5)^2+y^2=25$: centre $(5,0)$, radius $5$. The constant term is $0$, so $(0,0)$ satisfies the equation (equivalently, the centre is exactly $r=5$ from the origin).

7 $(x+3)^2+(y-2)^2=13-k$. A circle needs $13-k\gt0$, i.e. $k\lt13$. Radius $2$: $13-k=4$, $k=9$.

8 (a) $(x-1)^2+(y+2)^2=1+4-5=0$: the single point $(1,-2)$. (b) $(x+1)^2+(y+1)^2=1+1-3=-1$: no points at all.

9 Centre $(2,1)$, $r^2=4+1+11=16$. The distance squared from $(3,4)$ to the centre is $1+9=10\lt16$: inside. (Quick test: $9+16-12-8-11=-6\lt0$ means inside.)

III

Circles from conditions.

1 Find the circle through $(0,0)$, $(4,0)$ and $(0,6)$, and its centre and radius.
2 Find the circle through $(4,6)$, $(-2,-2)$ and $(5,-1)$.
3 Show that the point $(-3,5)$ lies on the circle of part 2. (The four points are concyclic.)
4 Find the circle through $(1,3)$ and $(5,-1)$ whose centre lies on the line $y=2x-5$.
5 Find the circles of radius $5$ with centre on the $y$-axis that pass through $(3,2)$.
6 Find the circle through the origin that meets the positive $x$-axis at $(6,0)$ and the negative $y$-axis at $(0,-4)$.
7 Find the circle with centre $(3,4)$ that passes through the centre of $x^2+y^2+4x-2y-4=0$.
8 Find the circles of radius $4$ that touch both coordinate axes and have their centres on the line $y=x$.

💡 Show solutions▶

1 Through $(0,0)$: $c=0$. $(4,0)$: $16+8g=0$, $g=-2$. $(0,6)$: $36+12f=0$, $f=-3$. So $x^2+y^2-4x-6y=0$: centre $(2,3)$, radius $\sqrt{13}$.

2 Substituting: $52+8g+12f+c=0$, $8-4g-4f+c=0$, $26+10g-2f+c=0$. Subtracting in pairs: $44+12g+16f=0$ (first minus second) and $18+14g+2f=0$ (third minus second), i.e. $3g+4f=-11$ and $7g+f=-9$; so $g=-1$, $f=-2$, $c=-20$. Answer $x^2+y^2-2x-4y-20=0$ (centre $(1,2)$, radius $5$).

3 $9+25+6-20-20=0$ ✓, so $(-3,5)$ is on the circle (its distance from $(1,2)$ is $\sqrt{16+9}=5$).

4 The centre $(h,k)$ is equidistant from the points: $(h-1)^2+(k-3)^2=(h-5)^2+(k+1)^2$ gives $8h-8k=16$, i.e. $h-k=2$. With $k=2h-5$: $h-2h+5=2$, $h=3$, $k=1$. $r^2=4+4=8$: $(x-3)^2+(y-1)^2=8$, i.e. $x^2+y^2-6x-2y+2=0$.

5 Centre $(0,k)$: $9+(2-k)^2=25$, $(2-k)^2=16$, $k=6$ or $k=-2$. Circles $x^2+(y-6)^2=25$ and $x^2+(y+2)^2=25$.

6 $c=0$, $36+12g=0$ gives $g=-3$, $16-8f=0$ gives $f=2$: $x^2+y^2-6x+4y=0$ (centre $(3,-2)$, radius $\sqrt{13}$).

7 The other centre is $(-2,1)$; $r^2=25+9=34$: $(x-3)^2+(y-4)^2=34$, i.e. $x^2+y^2-6x-8y-9=0$.

8 Touching both axes with radius $4$ puts the centre at $(\pm4,\pm4)$; on $y=x$ only $(4,4)$ and $(-4,-4)$: $(x-4)^2+(y-4)^2=16$ and $(x+4)^2+(y+4)^2=16$.

Cone sections and applications.

1 A double cone has semi-vertical angle $40^\circ$. A plane not through the vertex makes an angle of $40^\circ$ with the axis. Name the section.
2 A plane through the vertex of a double cone with $\alpha=25^\circ$ is perpendicular to the axis. What is the section?
3 For a double cone with $\alpha=30^\circ$, which angles $\beta$ (planes not through the vertex) give an ellipse? Which give a hyperbola?
4 A Ferris wheel has diameter $40\ \text{m}$ and its centre is $22\ \text{m}$ above the ground. Take the ground as the $x$-axis and the vertical line through the centre as the $y$-axis. Write the equation of the rim, and find the heights of the two points of the rim that are $12\ \text{m}$ horizontally from that vertical line.
5 An archaeologist finds a piece of a circular plate. Placing it on graph paper (units cm), three points of its rim are at $(0,0)$, $(16,0)$ and $(8,4)$. Find the centre and the diameter of the whole plate.
6 A mobile tower at $T(3,-2)$ (units km) covers everything within $5\ \text{km}$. Write the equation of the edge of its coverage and decide whether villages at $(0,1)$ and $(8,1)$ are covered.
7 The equation $2x^2+ky^2+6x-4y+1=0$ represents a circle. Find $k$, the centre and the radius.

💡 Show solutions▶

1 $\beta=\alpha$: a parabola.

2 Through the vertex with $\beta=90^\circ\gt\alpha$: a single point (the vertex).

3 Ellipse: $30^\circ\lt\beta\lt90^\circ$. Hyperbola: $0^\circ\le\beta\lt30^\circ$.

4 Centre $(0,22)$, radius $20$: $x^2+(y-22)^2=400$. At $x=\pm12$: $(y-22)^2=256$, $y=22\pm16$, i.e. $38\ \text{m}$ and $6\ \text{m}$ above the ground.

5 Through the origin: $c=0$. $(16,0)$: $256+32g=0$, $g=-8$. $(8,4)$: $64+16-128+8f=0$, $f=6$. Centre $(8,-6)$, $r=\sqrt{64+36}=10$: the plate's diameter is $20\ \text{cm}$.

6 $(x-3)^2+(y+2)^2=25$, i.e. $x^2+y^2-6x+4y-12=0$. $(0,1)$: $TP^2=9+9=18\lt25$, covered. $(8,1)$: $TP^2=25+9=34\gt25$, not covered.

7 A circle needs equal coefficients of $x^2$ and $y^2$: $k=2$. Dividing by $2$: $x^2+y^2+3x-2y+\tfrac12=0$, so $\left(x+\tfrac32\right)^2+(y-1)^2=\tfrac94+1-\tfrac12=\tfrac{11}{4}$: centre $\left(-\tfrac32,1\right)$, radius $\tfrac{\sqrt{11}}{2}$.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask for the centre and radius of a given circle (1–2 marks), the equation of a circle from its centre and a point or from the ends of a diameter (2 marks), and a circle through given points or with its centre on a given line (3 marks); case studies use fountains, tracks and wheels. JEE combines tangency with the axes, signs of $g$, $f$, $c$ and statements about a circle to be checked one by one.

E1. Board · 1 mark The centre and radius of $x^2+y^2+4x-6y-12=0$ are (A) $(2,-3)$, $5$ (B) $(-2,3)$, $5$ (C) $(-2,3)$, $25$ (D) $(-4,6)$, $5$

💡 Show solution▶

Idea $2g=4$, $2f=-6$, so $g=2$, $f=-3$, $c=-12$: centre $(-g,-f)=(-2,3)$, $r=\sqrt{4+9+12}=5$.

Trap (A) forgets to change the signs; (C) gives $r^2$; (D) uses the coefficients $4$ and $-6$ instead of half of them.

Answer (B).

E2. Board · 2 marks Find the equation of the circle with centre $(-3,1)$ that passes through the point $(1,-2)$.

💡 Show solution▶

Radius $r^2=(1+3)^2+(-2-1)^2=16+9=25$. (1 mark)

Answer $(x+3)^2+(y-1)^2=25$, i.e. $x^2+y^2+6x-2y-15=0$. (1 mark)

E3. Board · 3 marks Find the equation of the circle that passes through $(-1,2)$ and $(3,-2)$ and has its centre on the line $x+2y=4$.

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Equidistant Centre $(h,k)$: $(h+1)^2+(k-2)^2=(h-3)^2+(k+2)^2$. Expanding and cancelling $h^2$, $k^2$: $2h+1-4k+4=-6h+9+4k+4$, so $8h-8k=8$, i.e. $h-k=1$. (1 mark)

Centre With $h+2k=4$: subtracting, $3k=3$, so $k=1$, $h=2$. (1 mark)

Answer $r^2=(2+1)^2+(1-2)^2=10$: $(x-2)^2+(y-1)^2=10$, i.e. $x^2+y^2-4x-2y-5=0$. Check $(3,-2)$: $1+9=10$ ✓. (1 mark)

E4. Board · Case study The rim of a circular fountain in a city park is drawn on the park plan (units metres) as $x^2+y^2-8x-6y=0$. The park gate is at the origin and a straight path runs along the $x$-axis. (i) Find the centre and radius of the rim. (1 mark) (ii) Does the rim pass through the gate? (1 mark) (iii) Find the points where the path meets the rim, and the length of path that lies inside the fountain. (2 marks)

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(i) $(x-4)^2+(y-3)^2=16+9=25$: centre $(4,3)$, radius $5\ \text{m}$. (1 mark)

(ii) Yes: the constant term is $0$, so $(0,0)$ satisfies the equation (the gate is $\sqrt{16+9}=5\ \text{m}$ from the centre). (1 mark)

(iii) set-up Put $y=0$: $x^2-8x=0$, so $x=0$ or $x=8$. (1 mark)

Answer The path meets the rim at $(0,0)$ and $(8,0)$, so $8\ \text{m}$ of it lies inside the fountain. (1 mark)

E5. JEE Main The circle $x^2+y^2+2gx+2fy+c=0$ has radius $3$, touches both coordinate axes and has its centre in the second quadrant. Then $g+f+c$ equals (A) $9$ (B) $27$ (C) $-9$ (D) $3$

💡 Show solution▶

Idea Touching both axes means the centre is $3$ from each axis; in the second quadrant it is $(-3,3)$. So $-g=-3$, $-f=3$: $g=3$, $f=-3$.

$c$ $r^2=g^2+f^2-c$: $9=9+9-c$, so $c=9$. Then $g+f+c=3-3+9=9$.

Trap (B) comes from writing $c=g^2+f^2+r^2$; (C) from taking the centre's coordinates as $(g,f)$ and $c=-9$.

Answer (A).

⏱ Fast route A circle touching both axes with centre $(\pm r,\pm r)$ always has $c=r^2$ (here $9$), and $g+f=0$ in the second quadrant.

E6. JEE Advanced One or more options correct. For the circle $x^2+y^2-6x+8y=0$, which of the following are true? (A) it passes through the origin (B) its radius is $5$ (C) its centre is $(3,-4)$ (D) it touches the $x$-axis

💡 Show solution▶

Complete squares $(x-3)^2+(y+4)^2=9+16=25$: centre $(3,-4)$, radius $5$. So (B) and (C) are true.

(A) No constant term, so $(0,0)$ is on it: true.

(D) The centre is $4$ from the $x$-axis but the radius is $5\gt4$, so the circle cuts the $x$-axis (at $x=0$ and $x=6$): false. (It does not touch the $y$-axis either: $3\lt5$.)

Answer (A), (B), (C).

🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. How accurately can you find the size of a whole circular object — a plate, a wheel, a bangle — from a small broken piece of it?

Why it matters. Archaeologists at excavations such as Keezhadi in Tamil Nadu or Harappan sites in Gujarat find mostly small fragments of pots. From a piece of the rim they estimate the pot's original diameter, which tells them what the pot was used for and how many people it fed. Engineers do the same when they check a worn pipe or a railway wheel from a short arc. Three points fix a circle, so the general equation $x^2+y^2+2gx+2fy+c=0$ is all you need.

What to do.

Take a round object whose diameter you can measure directly (a steel plate, a bangle, a lid). Lay a short arc of its edge — no more than a quarter of the circle — on graph paper and trace it. Pretend the rest is missing.
Choose axes, and read the coordinates of three well-separated points on the traced arc.
Substitute them into $x^2+y^2+2gx+2fy+c=0$, solve for $g$, $f$, $c$ and compute the radius $\sqrt{g^2+f^2-c}$.
Also measure the chord $w$ joining the ends of your arc and the height $s$ of the arc above its mid-point, and use $r=\dfrac{w^2/4+s^2}{2s}$ (it comes from the right triangle formed by the radius, half the chord and $r-s$).
Compare both estimates with the true diameter. Repeat with a shorter arc.

Tools: Graph paper or a GeoGebra-style app with a photo of the arc, a ruler, a calculator or spreadsheet

Hand in: Your traced arc with the three points marked, the equations you solved, both estimates of the diameter, the percentage error of each, and a sentence on how the error changes as the arc gets shorter.

Going further. Read off five or six points instead of three. They will not lie exactly on one circle. Fit a circle to all of them by choosing $g$, $f$, $c$ to make the values of $x^2+y^2+2gx+2fy+c$ at your points as close to $0$ as possible (a spreadsheet solver can do this). Is the fitted diameter better?

Instructor note: errors of $1\ \text{mm}$ in reading the points matter far more for short arcs — the three points then lie almost on a line and the linear system is nearly singular, so $r$ swings wildly. A quarter arc usually gives the diameter within a few per cent. The sagitta formula is a good link to Class 10 geometry (perpendicular from the centre bisects a chord).

🧪 Self-check quiz

0/5 correct

Q1. The centre of the circle $(x-2)^2+(y+7)^2=9$ is

Q2. The radius of $x^2+y^2-6x+2y+1=0$ is

Q3. A plane that does not pass through the vertex of a double cone is parallel to one of its generators. The section is

Q4. Which equation represents a circle?

Q5. The circle on the segment joining $(0,0)$ and $(6,8)$ as diameter is

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