JEE Mains Maths · Conic Sections

Equation of a Circle

Reading a circle's centre and radius from any form of its equation, and building the equation from given conditions: points it passes through, lines it touches, intercepts it cuts, or a locus rule.

All Conic Sections notes JEE Mains Maths strategy

Why this matters

Forty-nine PYQs, the largest page in Conic Sections. Most are two steps: find the centre and radius from the data, then read off what is asked. The rest are loci that turn out to be circles. Six ideas cover all of them.

Concept 1 of 6: The general equation: centre, radius and when it is a circle

Every circle is

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

. Expand it and you get

x^2+y^2+2gx+2fy+c=0

with centre

(-g,-f)

and

r^2=g^2+f^2-c

. So a second-degree equation is a circle only when

x^2

and

y^2

carry equal coefficients and there is no

xy

term. Divide by the common coefficient before reading

g

f

and

c

Definition

Centre-radius form: $(x-h)^2+(y-k)^2=r^2$ .
General form: $x^2+y^2+2gx+2fy+c=0$ , centre $(-g,-f)$ , radius $\sqrt{g^2+f^2-c}$ .
Real circle: $g^2+f^2-c>0$ . If it is $0$ the circle is a single point; if negative there is no real circle.
Is it a circle? $ax^2+2hxy+by^2+\dots=0$ is a circle only if $a=b\neq0$ and $h=0$ .

Centre and radius of the general form

C=(-g,-f),\qquad r=\sqrt{g^2+f^2-c}

Worked example

Find the centre and radius of

2x^2+2y^2-8x+12y+6=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 27 July 2022 · Q79Moderate

Example 1 · Conic Sections · Equation of a Circle

If the circle

x^{2}+y^{2}- 2gx + 6y - 19c = 0,g,c \in R

passes through the point

(6,1)

and its centre lies on the line

x - 2cy = 8

, then the length of intercept made by the circle on

x

-axis is

Divide by the $x^2$ coefficient first

2x^2+2y^2-8x+12y+6=0

, reading

2g=-8

directly gives the wrong centre

(4,-6)

. The formulas

(-g,-f)

and

\sqrt{g^2+f^2-c}

hold only after the

x^2

and

y^2

coefficients are

1

The centre is $(-g,-f)$ , with the signs flipped

x^2+y^2-4x+6y=0

has

2g=-4

, so

g=-2

and the centre's

x

-coordinate is

+2

. Half the coefficient, then change the sign.

Drill 5 more on the general equation: centre, radius and when it is a circle

Concept 2 of 6: Position of a point, and nearest and farthest distances

Put a point

P(x_1,y_1)

into the left side of

S=x^2+y^2+2gx+2fy+c

. The number you get,

S_1

, equals

PC^2-r^2

. So its sign tells you where

P

is. The nearest and farthest points of the circle from

P

both lie on the line

PC

, at distances

|PC-r|

and

PC+r

Definition

$S_1<0$ : $P$ is inside. $S_1=0$ : on the circle. $S_1>0$ : outside.
Nearest distance from $P$ to the circle: $|PC-r|$ .
Farthest distance: $PC+r$ .
Both extreme points lie on the line through $P$ and the centre $C$ .

Power of a point

S_1=x_1^2+y_1^2+2gx_1+2fy_1+c=PC^2-r^2

Worked example

(1,2)

inside

x^2+y^2-4x-2y-4=0

? Find its nearest and farthest distances from the circle.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 12 · Q41Moderate

Example 2 · Conic Sections · Equation of a Circle

Let

r_{1}

and

r_{2}

be the radii of the largest and smallest circles, respectively, which pass through the point

( - 4,1)

and having their centres on the circumference of the circle

x^{2}+y^{2}+ 2x+ 4y- 4 = 0

. If

\frac{r_{1}}{r_{2}}=a+b\sqrt{2}

, then

a+b

is equal to:

For a point inside, the nearest distance is $r-PC$

Writing

PC-r

for an inside point gives a negative 'distance'. Use

|PC-r|

: it is

PC-r

outside and

r-PC

inside.

Drill 1 more on position of a point, and nearest and farthest distances

Concept 3 of 6: The diameter form and right angles

A

and

B

are the ends of a diameter, every other point

P

of the circle sees

AB

at a right angle, so

\vec{PA}\cdot\vec{PB}=0

. Written out, that is the diameter form. Read it backwards too: when a right angle stands on a fixed segment, its vertex lies on the circle with that segment as diameter. This is how four points are shown to be concyclic, and why the circumcentre of a right triangle is the midpoint of its hypotenuse.

Definition

Diameter form: ends $(x_1,y_1)$ , $(x_2,y_2)$ give $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$ .
Angle in a semicircle is $90^\circ$ .
Right triangle: circumcentre = midpoint of the hypotenuse, circumradius = half the hypotenuse.
If $x_1,x_2$ are roots of one quadratic and $y_1,y_2$ of another, the equation needs only their sums and products (Vieta). Do not solve for the roots.

Diameter form

(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0

Worked example

The points

(2,0)

(0,4)

(0,0)

and

(k,k)

k\neq0

, lie on one circle. Find

k

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q128Moderate

Example 3 · Conic Sections · Equation of a Circle

If the four distinct points

(4,6),( - 1,5),(0,0)

and (

k,3k

) lie on a circle of radius

r

, then

10k+r^{2}

is equal to.

Both points must be ends of ONE diameter

The diameter form through two points that are merely ON the circle gives a different, smaller circle. First check that the segment really is a diameter, for example by a right angle standing on it.

Drill 8 more on the diameter form and right angles

Concept 4 of 6: Touching the axes and cutting intercepts

To meet the

x

-axis put

y=0

x^2+2gx+c=0

. Its roots are the ends of the

x

-intercept, which has length

2\sqrt{g^2-c}

. The circle touches the

x

-axis when that length is zero, which happens exactly when the radius equals the centre's distance from the axis,

|k|

. The

y

-axis works the same way with

f

Definition

$x$ -intercept $=2\sqrt{g^2-c}$ ; $y$ -intercept $=2\sqrt{f^2-c}$ .
Touches the $x$ -axis: $r=|k|$ (equivalently $g^2=c$ ). Touches the $y$ -axis: $r=|h|$ .
Touches both axes: centre $(\pm r,\pm r)$ , signs chosen by the quadrant.
Meets neither axis: $r<|h|$ and $r<|k|$ .
A chord at distance $d$ from the centre has length $2\sqrt{r^2-d^2}$ ; an intercept is the case where the chord is an axis.

Intercepts on the axes

\ell_x=2\sqrt{g^2-c},\qquad \ell_y=2\sqrt{f^2-c}

Worked example

A circle touches the

x

-axis at

(3,0)

, lies above it and cuts an intercept of

8

on the

y

-axis. Find its equation.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 17 · Q77Moderate

Example 4 · Conic Sections · Equation of a Circle

Consider a circle

C

which touches the

y

-axis at

(0,6)

and cuts off an intercept

6\sqrt{5}

on the x-axis. Then the radius of the circle

C

is equal to:

Touching the $x$ -axis fixes $r=|k|$ , not $|h|$

The radius to the point of contact is perpendicular to the axis, so it runs vertically: its length is the centre's

y

-coordinate. Mixing up

h

and

k

here is the usual slip.

Drill 5 more on touching the axes and cutting intercepts

Concept 5 of 6: Building a circle from conditions

A circle has three unknowns,

h

k

r

, so it needs three conditions. Each condition gives one equation: passing through a point (substitute it), centre on a line (substitute the centre), touching a line (distance from the centre equals

r

), touching a curve at a known point (the centre lies on the normal there). Pick the form in which the given conditions are easiest to write.

Definition

Through $A$ and $B$ : the centre is on the perpendicular bisector of $AB$ .
Touches a line $L$ : the distance from the centre to $L$ equals $r$ .
Touches $L$ at $P$ : the centre is on the normal to $L$ at $P$ , at distance $r$ .
Touches two parallel lines: $2r$ is the gap between them and the centre is on the midway line.
Touches two crossing lines: the centre is on a bisector of the angle between them.

Distance from the centre to a tangent line

\frac{|ah+bk+c|}{\sqrt{a^2+b^2}}=r

Worked example

A circle passes through

(1,0)

and

(5,0)

and touches the

y

-axis. Find it.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 June 2022 · Q72Moderate

Example 5 · Conic Sections · Equation of a Circle

Let

C

be a circle passing through the points

A(2, - 1)

and

B(3,4)

. The line segment

AB

is not a diameter of

C

. If

r

is the radius of

C

and its centre lies on the circle

(x - 5)^{2}+ (y - 1)^{2}=\frac{13}{2}

, then

r^{2}

is equal to :

Two parallel tangents give the DIAMETER, not the radius

The gap between two parallel tangents spans the whole circle, so

r

is half of it. Taking the gap as

r

doubles the radius.

A distance condition gives two signs

|ah+bk+c|=r\sqrt{a^2+b^2}

splits into two cases. Keep both until a stated condition (a quadrant, 'below the axis') rules one out.

Drill 13 more on building a circle from conditions

Concept 6 of 6: Loci that turn out to be circles

Call the moving point

(h,k)

, turn the condition into an equation, simplify, and rename

h,k

x,y

. Several conditions always give circles: a fixed ratio of distances from two points (other than

1

), a fixed sum of squared distances from fixed points, and a point that divides a segment whose other end moves on a circle. When the point is given by a parameter, remove it with

\cos^2\theta+\sin^2\theta=1

Definition

$PA=\lambda\,PB$ with $\lambda\neq1$ : a circle. With $\lambda=1$ it is the perpendicular bisector, a line.
$\sum PA_i^2=$ constant: a circle centred at the centroid of the fixed points.
$A$ fixed, $B$ moving on a circle of centre $O$ , radius $r$ ; $P$ divides $AB$ as $m:n$ from $A$ : $P$ moves on a circle of centre $\frac{nA+mO}{m+n}$ and radius $\frac{m}{m+n}r$ .
Parameter $\theta$ : isolate $\cos\theta$ and $\sin\theta$ , then square and add.

A dividing point whose far end moves on a circle

P=\frac{nA+mB}{m+n}\ \Rightarrow\ \text{radius}=\frac{m}{m+n}\,r

Worked example

Find the locus of

P

with

PA=2\,PB

, where

A=(0,0)

and

B=(3,0)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 2 · Q87Moderate

Example 6 · Conic Sections · Equation of a Circle

Let a point

P

be such that its distance from the point

(5,0)

is thrice the distance of

P

from the point

( - 5,0)

. If the locus of the point

P

is a circle of radius

r

, then

4r^{2}

is equal to

The radius scales by the MOVING end's share

P

divides

AB

m:n

from the fixed point

A

P

\frac{m}{m+n}

of the way to

B

, so its circle has radius

\frac{m}{m+n}r

. Using

\frac{n}{m+n}

swaps the shares.

Equal distances give a line, not a circle

The ratio rule gives a circle only for

\lambda\neq1

. When

PA=PB

, the squared terms cancel and the locus is the perpendicular bisector.

Drill 11 more on loci that turn out to be circles

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (6)

The general equation: centre, radius and when it is a circle
Centre and radius of the general form
$C=(-g,-f),\qquad r=\sqrt{g^2+f^2-c}$
Position of a point, and nearest and farthest distances
Power of a point
$S_1=x_1^2+y_1^2+2gx_1+2fy_1+c=PC^2-r^2$
The diameter form and right angles
Diameter form
$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$
Touching the axes and cutting intercepts
Intercepts on the axes
$\ell_x=2\sqrt{g^2-c},\qquad \ell_y=2\sqrt{f^2-c}$
Building a circle from conditions
Distance from the centre to a tangent line
$\frac{|ah+bk+c|}{\sqrt{a^2+b^2}}=r$
Loci that turn out to be circles
A dividing point whose far end moves on a circle
$P=\frac{nA+mB}{m+n}\ \Rightarrow\ \text{radius}=\frac{m}{m+n}\,r$

Watch out for (9)

Divide by the $x^2$ coefficient first
→ The general equation: centre, radius and when it is a circle
The centre is $(-g,-f)$ , with the signs flipped
→ The general equation: centre, radius and when it is a circle
For a point inside, the nearest distance is $r-PC$
→ Position of a point, and nearest and farthest distances
Both points must be ends of ONE diameter
→ The diameter form and right angles
Touching the $x$ -axis fixes $r=|k|$ , not $|h|$
→ Touching the axes and cutting intercepts
Two parallel tangents give the DIAMETER, not the radius
→ Building a circle from conditions
A distance condition gives two signs
→ Building a circle from conditions
The radius scales by the MOVING end's share
→ Loci that turn out to be circles
Equal distances give a line, not a circle
→ Loci that turn out to be circles

Test yourself on Conic Sections

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