JEE Mains Maths · Conic Sections

Chords and Tangents of a Circle

How a line meets one circle: the length of the chord it cuts, the chord with a given midpoint, when a line is a tangent, and what the two tangents from an outside point make: their length, the chord of contact, the angle and the areas.

All Conic Sections notes JEE Mains Maths strategy

Why this matters

Forty-six PYQs. Nearly every one reduces to one fact: the distance from the centre to a line, compared with the radius. Less than the radius gives a chord, equal gives a tangent. Five ideas cover the page.

Concept 1 of 5: Chord length from the distance to the centre

Drop a perpendicular from the centre to the chord. It bisects the chord, so half the chord, the perpendicular

d

and the radius make a right triangle. That gives the chord length at once, without finding the two ends. The same picture answers 'which chord through a point is longest or shortest': the longest is the diameter, the shortest is perpendicular to the line joining the point to the centre.

Definition

Chord at distance $d$ from the centre: length $2\sqrt{r^2-d^2}$ .
The perpendicular from the centre bisects every chord.
$d<r$ : the line cuts the circle; $d=r$ : it touches; $d>r$ : it misses.
Through a point $P$ inside: the longest chord is the diameter; the shortest (and the one farthest from the centre) is perpendicular to $CP$ .

Chord length

\ell=2\sqrt{r^2-d^2}

Worked example

Find the length of the chord that

3x+4y=10

cuts on

x^2+y^2=25

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 14 June 2022 · Q177Moderate

Example 1 · Conic Sections · Chords and Tangents of a Circle

Let a circle

C:(x - h)^{2}+ (y - k)^{2}=r^{2},k > 0

, touch the

x

-axis at

(1,0)

. If the line

x + y = 0

intersects the circle

C

P

and

Q

such that the length of the chord

PQ

is 2 , then the value of

h + k + r

is equal to

Use the distance from the CENTRE, not from the origin

For a circle not centred at the origin, the chord length needs the perpendicular distance from

(-g,-f)

to the line. Measuring from

(0,0)

out of habit gives a wrong

d

Drill 11 more on chord length from the distance to the centre

Concept 2 of 5: Chords by their midpoint, or by the angle they subtend

The chord with a given midpoint

M

is perpendicular to

CM

, so its equation is fixed. The shortcut

T=S_1

writes it in one line. When a chord through a fixed point varies, its midpoints

M

all see

CP

at a right angle, so they lie on the circle with

CP

as diameter. And a chord that subtends a right angle at the origin is found by homogenising: combine the circle and the line into one equation of degree two and ask that its two lines through the origin be perpendicular.

Definition

$T=xx_1+yy_1+g(x+x_1)+f(y+y_1)+c$ and $S_1$ is the circle's value at $(x_1,y_1)$ .
Chord with midpoint $(x_1,y_1)$ : $T=S_1$ .
Midpoints of chords through $P$ : the circle on $CP$ as diameter (the part inside the given circle).
Chord subtending $90^\circ$ at the origin: write the chord as $lx+my=1$ , make the circle's equation homogeneous with it, and set (coefficient of $x^2$ ) + (coefficient of $y^2$ ) $=0$ .

Chord with a given midpoint

T=S_1

Worked example

Find the chord of

x^2+y^2-4x-6y-12=0

whose midpoint is

(1,1)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q139Moderate

Example 2 · Conic Sections · Chords and Tangents of a Circle

Let a circle

C

pass through the points

(4,2)

and

(0

, 2 ), and its centre lie on

3x + 2y + 2 = 0

. Then the length of the chord, of the circle

C

, whose midpoint is

(1,2)

, is:

$T=S_1$ is not $T=0$

T=0

is the tangent (or polar) at

(x_1,y_1)

. The chord with midpoint

(x_1,y_1)

T=S_1

. Mixing them gives a line through the wrong point.

Drill 5 more on chords by their midpoint, or by the angle they subtend

Concept 3 of 5: When a line is a tangent, and the tangent at a point

A line is a tangent exactly when its distance from the centre equals the radius. The radius to the point of contact is perpendicular to the tangent, so the normal at any point passes through the centre. That is why two normals meet at the centre, and why a circle touching three lines is the incircle (or an excircle) of the triangle they form.

Definition

Tangent test: distance from the centre $=r$ .
Tangent at $(x_1,y_1)$ on the circle: $T=0$ , i.e. $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$ .
Slope form for $x^2+y^2=a^2$ : $y=mx\pm a\sqrt{1+m^2}$ .
Normal: passes through the centre, so two normals meet at the centre.

Tangent of slope m to x² + y² = a²

y=mx\pm a\sqrt{1+m^2}

Worked example

Find the tangents of slope

2

x^2+y^2=9

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 7 · Q74Moderate

Example 3 · Conic Sections · Chords and Tangents of a Circle

The line

2x-y+ 1 = 0

is a tangent to the circle at the point

(2,5)

and the centre of the circle lies on

x- 2y= 4

. Then, the radius of the circle is:

The slope form needs the circle centred at the origin

y=mx\pm a\sqrt{1+m^2}

is for

x^2+y^2=a^2

. For any other circle, use the distance test from the actual centre, or shift the origin first.

Drill 10 more on when a line is a tangent, and the tangent at a point

Concept 4 of 5: Tangents from an outside point: length, chord of contact, angle, area

From an outside point

P

, two tangents touch the circle at

A

and

B

. The picture is symmetric about

PC

: the two tangents are equal,

PA\perp CA

, and

AB

is perpendicular to

PC

. Every quantity comes from the right triangle

PAC

with legs

L

(tangent length) and

r

. The line

AB

, the chord of contact, has the same equation

T=0

as a tangent. Here

(x_1,y_1)

is the outside point, so it is the polar of

P

Definition

Tangent length: $L=\sqrt{S_1}$ , where $S_1=PC^2-r^2$ .
Chord of contact (polar of $P$ ): $T=0$ .
Angle between the tangents: $2\tan^{-1}\frac{r}{L}$ .
Chord of contact length: $\frac{2rL}{\sqrt{r^2+L^2}}$ .
Area of triangle $PAB$ : $\frac{rL^3}{r^2+L^2}$ . Area of quadrilateral $PACB$ : $rL$ .
Power of a point: any line through $P$ meeting the circle at $R$ , $S$ has $PR\cdot PS=S_1$ .

Tangent length and area of triangle PAB

L=\sqrt{S_1},\qquad [PAB]=\frac{rL^3}{r^2+L^2}

Worked example

From

P(4,3)

tangents are drawn to

x^2+y^2=9

. Find the tangent length, the chord of contact and the angle between the tangents.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 28 July 2022 · Q163Moderate

Example 4 · Conic Sections · Chords and Tangents of a Circle

Let the tangents at two points

A

and

B

on the circle

x^{2}+y^{2}- 4x + 3 = 0

meet at origin

O(0,0)

. Then the area of the triangle of

OAB

The angle is TWICE $\tan^{-1}\frac{r}{L}$

\tan^{-1}\frac{r}{L}

is the angle between one tangent and

PC

. The angle between the two tangents is double it.

Triangle $PAB$ is not triangle $PAC$

[PAC]=\frac12rL

is half the quadrilateral

PACB

. The triangle cut off by the chord of contact is smaller:

\frac{rL^3}{r^2+L^2}

Drill 13 more on tangents from an outside point: length, chord of contact, angle, area

Concept 5 of 5: Parametric points and largest and smallest values

Every point of

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

(h+r\cos\theta,\ k+r\sin\theta)

. That turns 'the largest value of an expression over the circle' into a one-variable problem. For a distance from a fixed point, you do not even need the parameter: the extremes lie on the line through the centre.

Definition

Parametric point: $(h+r\cos\theta,\ k+r\sin\theta)$ .
$a\cos\theta+b\sin\theta$ ranges over $[-\sqrt{a^2+b^2},\ \sqrt{a^2+b^2}]$ .
Distance from a fixed point $Q$ to the circle: between $|QC-r|$ and $QC+r$ .

Parametric point

(h+r\cos\theta,\ k+r\sin\theta)

Worked example

Find the largest and smallest values of

x^2+y^2

for points on

(x-3)^2+(y-4)^2=1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 5 Apr 2024 · Q72Moderate

Example 5 · Conic Sections · Chords and Tangents of a Circle

Let a circle

C

of radius 1 and closer to the origin be such that the lines passing through the point

(3,2)

and parallel to the coordinate axes touch it. Then the shortest distance of the circle

C

from the point

(5,5)

is:

Find the centre before measuring

Extremes of distance run along the line through the CENTRE. Working from a point on the circle that merely looks nearest gives the wrong value.

Drill 2 more on parametric points and largest and smallest values

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 (5)

Chord length from the distance to the centre
Chord length
$\ell=2\sqrt{r^2-d^2}$
Chords by their midpoint, or by the angle they subtend
Chord with a given midpoint
$T=S_1$
When a line is a tangent, and the tangent at a point
Tangent of slope m to x² + y² = a²
$y=mx\pm a\sqrt{1+m^2}$
Tangents from an outside point: length, chord of contact, angle, area
Tangent length and area of triangle PAB
$L=\sqrt{S_1},\qquad [PAB]=\frac{rL^3}{r^2+L^2}$
Parametric points and largest and smallest values
Parametric point
$(h+r\cos\theta,\ k+r\sin\theta)$

Watch out for (6)

Use the distance from the CENTRE, not from the origin
→ Chord length from the distance to the centre
$T=S_1$ is not $T=0$
→ Chords by their midpoint, or by the angle they subtend
The slope form needs the circle centred at the origin
→ When a line is a tangent, and the tangent at a point
The angle is TWICE $\tan^{-1}\frac{r}{L}$
→ Tangents from an outside point: length, chord of contact, angle, area
Triangle $PAB$ is not triangle $PAC$
→ Tangents from an outside point: length, chord of contact, angle, area
Find the centre before measuring
→ Parametric points and largest and smallest values

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