JEE Mains Maths · Conic Sections

Two Circles and Families of Circles

How two circles sit relative to each other and how many common tangents they have, their common chord, the circles through the points where two curves meet, and the image of a circle in a line.

All Conic Sections notes JEE Mains Maths strategy

Why this matters

Twenty-six PYQs. More than half are one comparison: the distance between the centres against the sum and the difference of the radii. The rest subtract one equation from the other, or reflect a centre. Four ideas cover them.

Concept 1 of 4: Relative position and the number of common tangents

Everything depends on

d

, the distance between the centres, set against

r_1+r_2

and

|r_1-r_2|

. Far apart, the circles have four common tangents. As they come closer, they touch from outside (three), cross (two), touch from inside (one), and finally one sits inside the other (none). When two circles touch, the point of contact lies on the line of centres and splits it in the ratio of the radii.

Definition

$d>r_1+r_2$ : separate, $4$ common tangents.
$d=r_1+r_2$ : touch externally, $3$ .
$|r_1-r_2|<d<r_1+r_2$ : meet at two points, $2$ .
$d=|r_1-r_2|$ : touch internally, $1$ .
$d<|r_1-r_2|$ : one inside the other, $0$ .
Point of contact: divides $C_1C_2$ as $r_1:r_2$ , internally for an outside touch, externally for an inside touch.
Least distance between separate circles: $d-r_1-r_2$ .

Two circles meet in two points when

|r_1-r_2|<C_1C_2<r_1+r_2

Worked example

How many common tangents do

x^2+y^2=4

and

x^2+y^2-6x-8y+16=0

have?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 15 Apr 2023 · Q72Moderate

Example 1 · Conic Sections · Two Circles and Families of Circles

The number of common tangents, to the circles

x^{2}+y^{2}- 18x - 15y + 131 = 0

and

x^{2}+y^{2}- 6x - 6y - 7 = 0

, is :

Two points needs BOTH inequalities

d<r_1+r_2

alone allows one circle to sit inside the other. Two intersection points needs

|r_1-r_2|<d

as well.

Drill 14 more on relative position and the number of common tangents

Concept 2 of 4: The common chord, and a diameter that is a chord of another circle

Subtract the two circle equations (both with

x^2+y^2

coefficient

1

). The squared terms cancel and a line is left. Its points satisfy both equations, so it passes through both intersection points: it is the common chord. For its length, find its distance from either centre. A related picture: when a diameter of one circle is a chord of another, the first centre is the chord's midpoint, so a right triangle links the two radii and the distance between the centres.

Definition

Common chord: $S_1-S_2=0$ .
Its length: $2\sqrt{r_1^2-p_1^2}$ , where $p_1$ is its distance from $C_1$ .
The common chord is perpendicular to the line of centres.
A diameter of circle 1 is a chord of circle 2: $r_2^2=r_1^2+C_1C_2^2$ .

Common chord

S_1-S_2=0

Worked example

Find the common chord of

x^2+y^2=25

and

x^2+y^2-8x+7=0

, and its length.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 5 Apr 2024 · Q162Moderate

Example 2 · Conic Sections · Two Circles and Families of Circles

Let the circle

C_{1}:x^{2}+y^{2}- 2(x + y) + 1 = 0

and

C_{2}

be a circle having centre at

( - 1,0)

and radius 2. If the line of the common chord of

C_{1}

and

C_{2}

intersects the

y

-axis at the point

P

, then the square of the distance of

P

from the centre of

C_{1}

is:

Make the $x^2$ coefficients equal before subtracting

S_1-S_2

gives a line only when both equations have

x^2+y^2

with coefficient

1

. Divide out first, or the squared terms survive.

Drill 4 more on the common chord, and a diameter that is a chord of another circle

Concept 3 of 4: Circles through the meeting points of two curves

S=0

and

L=0

both hold at a point, so does

S+\lambda L=0

for every

\lambda

. So

S+\lambda L=0

is a whole family of circles through the points where the circle and line meet. One more condition fixes

\lambda

. The same idea works for two circles,

S_1+\lambda S_2=0

, and for two conics: choose

\lambda

so that the

x^2

and

y^2

coefficients match, and the combination is a circle through their four meeting points.

Definition

Circle and line: $S+\lambda L=0$ .
Two circles: $S_1+\lambda S_2=0$ , $\lambda\neq-1$ ( $\lambda=-1$ gives the common chord).
Two conics: pick $\lambda$ so the combination has equal $x^2$ , $y^2$ coefficients and no $xy$ term; it is then a circle through all their common points.
A circle touching $L$ at $P$ : $(x-x_1)^2+(y-y_1)^2+\lambda L=0$ .

Family through a circle and a line

S+\lambda L=0

Worked example

Find the circle through the points where

x+y=1

meets

x^2+y^2=4

that also passes through

(1,1)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 19 · Q79Moderate

Example 3 · Conic Sections · Two Circles and Families of Circles

A circle

C

touches the line

x = 2y

at the point

(2,1)

and intersects the circle

C_{1}:x^{2}+y^{2}+ 2y- 5 = 0

at two points

P

and

Q

such that

PQ

is a diameter of

C_{1}

. Then the diameter of

C

is:

$\lambda=-1$ is not a circle

S_1+\lambda S_2=0

the

x^2+y^2

terms cancel at

\lambda=-1

, leaving the common chord. Exclude it when a circle is wanted.

Drill 1 more on circles through the meeting points of two curves

Concept 4 of 4: The image of a circle in a line

A reflection moves every point but keeps distances. So the image of a circle is a circle with the same radius, centred at the image of the centre. Only one point, the centre, has to be reflected.

Definition

Image radius = original radius.
Image centre = reflection of the original centre.
Reflection of $(x_0,y_0)$ in $ax+by+c=0$ : $(x_0,y_0)-\frac{2(ax_0+by_0+c)}{a^2+b^2}(a,b)$ .
In $y=x$ : swap the coordinates.

Reflection of a point in a line

(x',y')=(x_0,y_0)-\frac{2(ax_0+by_0+c)}{a^2+b^2}\,(a,b)

Worked example

Find the image of

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

y=x

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 July 2022 · Q90Moderate

Example 4 · Conic Sections · Two Circles and Families of Circles

Let the mirror image of a circle

c_{1}:x^{2}+y^{2}- 2x -

6y + \alpha = 0

in line

y = x + 1

c_{2}:5x^{2}+ 5y^{2}+ 10gx

+10 fy

+ 38 = 0

. If

r

is the radius of circle

c_{2}

, then

\alpha + 6r^{2}

is equal to

Equal radii give a second equation

When the image circle is given with unknown coefficients, the equal radii are a condition too:

r_1^2=r_2^2

often fixes the last constant. Reflecting only the centre and stopping leaves it unused.

Drill 3 more on the image of a circle in a line

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 (4)

Relative position and the number of common tangents
Two circles meet in two points when
$|r_1-r_2|<C_1C_2<r_1+r_2$
The common chord, and a diameter that is a chord of another circle
Common chord
$S_1-S_2=0$
Circles through the meeting points of two curves
Family through a circle and a line
$S+\lambda L=0$
The image of a circle in a line
Reflection of a point in a line
$(x',y')=(x_0,y_0)-\frac{2(ax_0+by_0+c)}{a^2+b^2}\,(a,b)$

Watch out for (4)

Two points needs BOTH inequalities
→ Relative position and the number of common tangents
Make the $x^2$ coefficients equal before subtracting
→ The common chord, and a diameter that is a chord of another circle
$\lambda=-1$ is not a circle
→ Circles through the meeting points of two curves
Equal radii give a second equation
→ The image of a circle in a line

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