Circle: Chord, Tangent And Normal

Q: What is the equation of a tangent to a circle at a given point?

For x^2 + y^2 = a^2 the tangent at (x_1, y_1) is xx_1 + yy_1 = a^2. For the general circle x^2 + y^2 + 2gx + 2fy + c = 0, it is xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0, often written as T = 0.

Q: How do you find the length of a tangent from an external point to a circle?

Compute L = \sqrt{S_1} where S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c evaluated at the external point (x_1, y_1). S_1 must be positive; if it isn't, the point is inside or on the circle and no tangent exists.

Q: What is the equation of the pair of tangents from an external point?

SS_1 = T^2, where S is the circle expression, S_1 is its value at the external point (a number), and T is xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c. This gives a second-degree equation representing the two tangent lines together.

Q: What is the chord of contact of a circle?

When two tangents are drawn from an external point P(x_1, y_1) to a circle, the line joining the two points of contact is the chord of contact. Its equation is T = 0, the same as the polar of P.

Q: What is the director circle and what is its equation?

The locus of points from which two perpendicular tangents can be drawn to a circle is its director circle. It is concentric with the given circle and has radius \sqrt{2} times as large. For x^2 + y^2 = a^2, the director circle is x^2 + y^2 = 2a^2.

Q: What is the difference between a pole, polar and chord of contact?

The polar of a point P with respect to a circle is the line whose pole is P; equation is T = 0. If P is external, the polar coincides with the chord of contact of the two tangents from P. If P is on the circle, the polar coincides with the tangent at P.

Q: How do you find the equation of a chord with a given midpoint?

Use T = S_1, where T is the tangent-form expression xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c evaluated symbolically, and S_1 is the value of the circle expression at the midpoint (x_1, y_1).

Q: What is the length of a chord cut off from a circle by a line?

If the perpendicular distance from the centre to the line is p and the radius is r, the chord length is 2\sqrt{r^2 - p^2}. This works because the perpendicular from the centre bisects the chord.

Q: Why does the normal to a circle always pass through the centre?

The tangent at any point P on a circle is perpendicular to the radius CP. The normal at P is the line perpendicular to the tangent through P, which is exactly the line along the radius. Hence every normal to a circle passes through the centre.

MathsCirclesFor JEE aspirants

The chord, tangent and normal of a circle are the three key lines that connect every JEE problem on a single circle. A tangent to the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ at the point $(x_{1}, y_{1})$ is $x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0$ (the $T = 0$ form). The normal at the same point is the line joining $(x_{1}, y_{1})$ to the centre $(- g, - f)$ . This concept covers line-circle intersection, tangent and normal in every form, length of tangent, pair of tangents, chord of contact, director circle, pole and polar, and chord with a given midpoint.

Key Formulas - Quick Reference

Line $y = m x + c$ is tangent to $x^{2} + y^{2} = a^{2}$ iff $c^{2} = a^{2} (1 + m^{2})$ ; tangent line: $y = m x \pm a 1 + m^{2}$
Tangent at $(x_{1}, y_{1})$ on $x^{2} + y^{2} = a^{2}$ : $x x_{1} + y y_{1} = a^{2}$
Tangent at $(x_{1}, y_{1})$ on general circle: $x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0$ (i.e. $T = 0$ )
Parametric tangent at $(a cos α, a sin α)$ on $x^{2} + y^{2} = a^{2}$ : $x cos α + y sin α = a$
Normal at $(x_{1}, y_{1})$ on general circle: $y - y_{1} = \frac{y _{1} + f}{x _{1} + g} (x - x_{1})$
Length of tangent from $(x_{1}, y_{1})$ : $L = S_{1}$ where $S_{1} = x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c$
Pair of tangents from external point: $S S_{1} = T^{2}$
Chord of contact from external point $(x_{1}, y_{1})$ : $T = 0$
Chord with midpoint $(x_{1}, y_{1})$ : $T = S_{1}$
Length of chord cut by a line at perpendicular distance $p$ from centre: $2 r^{2} - p^{2}$
Director circle of $x^{2} + y^{2} = a^{2}$ : $x^{2} + y^{2} = 2 a^{2}$ (radius $a 2$ , concentric)
Polar of $(x_{1}, y_{1})$ w.r.t. general circle: $T = 0$ (same equation form as tangent)

1. Line and a Circle

Let $L = 0$ be a line and $S = 0$ be a circle of radius $r$ . Let $p$ be the perpendicular distance from the centre of the circle to the line. Then

$p > r$ - the line does not meet the circle
$p = r$ - the line touches the circle (tangent)
$p < r$ - the line is a secant of the circle (cuts it at two points)
$p = 0$ - the line passes through the centre, so the chord is a diameter

Figure 1: A line at perpendicular distance

p

from the centre. If

p > r

the line misses the circle, if

p = r

it touches (tangent), and if

p < r

it is a secant.

If the line is $y = m x + c$ and the circle is $x^{2} + y^{2} = a^{2}$ , the same three cases become

$c^{2} < a^{2} (1 + m^{2})$ - secant
$c^{2} = a^{2} (1 + m^{2})$ - tangent
$c^{2} > a^{2} (1 + m^{2})$ - the line does not meet the circle

These come from substituting $y = m x + c$ into the circle equation and demanding the discriminant of the resulting quadratic be positive, zero or negative.

Length of a chord cut by a line

If a line at perpendicular distance $p$ from the centre of a circle of radius $r$ cuts the circle in a chord $A B$ , then

A B = 2 r^{2} - p^{2} .

Figure 2: The perpendicular from the centre bisects the chord. In right triangle

C M A

C A = r

C M = p

and

A M = r^{2} - p^{2}

, so

A B = 2 r^{2} - p^{2}

The perpendicular from the centre bisects the chord, so in the right triangle $△ C M A$ , $C A = r$ (hypotenuse), $C M = p$ (one leg), and $A M = r^{2} - p^{2}$ (the other). The chord is $2 A M$ .

Solved Example 1

For what value of

c

will the line

y = 2 x + c

be a tangent to the circle

x^{2} + y^{2} = 5

Solution:

Rewrite the line as $2 x - y + c = 0$ . Perpendicular distance from origin (centre) $= \frac{c}{4 + 1} = \frac{∣ c ∣}{5} .$

For tangency, this equals the radius $5$ : $\frac{∣ c ∣}{5} = 5 \Rightarrow c = \pm 5.$

Solved Example 2

Find the length of the chord that the line

3 x + 4 y - 10 = 0

cuts from the circle

x^{2} + y^{2} = 25

Solution:

Centre $(0, 0)$ , radius $r = 5$ . Perpendicular distance from $(0, 0)$ to the line:

$p = \frac{∣3 ( 0 ) + 4 ( 0 ) - 10∣}{9 + 16} = \frac{10}{5} = 2.$

Chord length $= 2 r^{2} - p^{2} = 2 25 - 4 = 221 .$

2. Tangent to a Circle

A tangent to a circle is a line that touches it at exactly one point (the point of contact). We derive the equation of a tangent in three convenient forms.

Figure 3: The tangent at

P (x_{1}, y_{1})

is perpendicular to the radius

C P

. On

x^{2} + y^{2} = a^{2}

its equation is

x x_{1} + y y_{1} = a^{2}

(a) Point form of tangent

Circle $x^{2} + y^{2} = a^{2}$ : the tangent at $(x_{1}, y_{1})$ on the circle is

x x_{1} + y y_{1} = a^{2} .

General circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ : the tangent at $(x_{1}, y_{1})$ on the circle is

x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0.

The general $T$ rule. To write the tangent to any second-degree curve at a point

(x_{1}, y_{1})

on it, start from the curve expression

S

and make the replacements below. The constant term is left untouched. The equation you obtain is written

T = 0

, and the same

T

reappears in almost every other result on this page.

In $S$ , replace	by
$x^{2}$	$x x_{1}$
$y^{2}$	$y y_{1}$
$x$	$\frac{x + x _{1}}{2}$
$y$	$\frac{y + y _{1}}{2}$
$x y$	$\frac{x _{1} y + x y _{1}}{2}$
constant term	unchanged

(b) Slope form of tangent

The line $y = m x + c$ is tangent to $x^{2} + y^{2} = a^{2}$ if and only if $c^{2} = a^{2} (1 + m^{2})$ . So for any slope $m$ the two tangents are

y = m x \pm a 1 + m^{2},

and their points of contact are $(\mp \frac{a ^{2} m}{c}, \pm \frac{a ^{2}}{c})$ where $c = \pm a 1 + m^{2}$ .

(c) Parametric form of tangent

The tangent to $x^{2} + y^{2} = a^{2}$ at the parametric point $(a cos α, a sin α)$ is

x cos α + y sin α = a .

Point of intersection of two parametric tangents. The tangents at

P (α)

and

Q (β)

x^{2} + y^{2} = a^{2}

meet at

(\frac{a cos \frac{α + β}{2}}{cos \frac{α - β}{2}}, \frac{a sin \frac{α + β}{2}}{cos \frac{α - β}{2}})

Solved Example 3

Find the equation of the tangent to the circle

x^{2} + y^{2} - 30 x + 6 y + 109 = 0

(4, - 1)

Solution:

Apply $T = 0$ with $x_{1} = 4, y_{1} = - 1$ , $g = - 15, f = 3, c = 109$ :

$4 x + (- y) - 15 (x + 4) + 3 (y - 1) + 109 = 0$

$\Rightarrow 4 x - y - 15 x - 60 + 3 y - 3 + 109 = 0$

$\Rightarrow - 11 x + 2 y + 46 = 0$ , i.e. $11 x - 2 y - 46 = 0.$

Solved Example 4

Find the equations of tangents to the circle

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

which are parallel to the line

4 x + 3 y + 5 = 0

Solution:

Centre $(3, - 2)$ , radius $9 + 4 + 12 = 5$ .

Any line parallel to $4 x + 3 y + 5 = 0$ has the form $4 x + 3 y + k = 0$ .

For tangency, distance from centre $= 5$ :

$\frac{∣4 ( 3 ) + 3 ( - 2 ) + k ∣}{16 + 9} = 5 \Rightarrow ∣6 + k ∣ = 25 \Rightarrow k = 19$ or $k = - 31.$

Required tangents: $4 x + 3 y + 19 = 0$ and $4 x + 3 y - 31 = 0.$

3. Normal to a Circle

A normal to a circle at a point $P$ is the line through $P$ perpendicular to the tangent at $P$ . Since the radius $C P$ is perpendicular to the tangent, every normal passes through the centre of the circle. So the normal at $P (x_{1}, y_{1})$ is simply the line joining $P$ to the centre.

For the general circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ with centre $(- g, - f)$ , the normal at $(x_{1}, y_{1})$ is

y - y_{1} = \frac{y _{1} + f}{x _{1} + g} (x - x_{1}) .

Solved Example 5

Find the equation of the normal to the circle

x^{2} + y^{2} - 5 x + 2 y - 48 = 0

(5, 6)

Solution:

Centre $= (\frac{5}{2}, - 1)$ . Slope of the line joining centre to $(5, 6)$ :

$m = \frac{6 - ( - 1 )}{5 - 5/2} = \frac{7}{5/2} = \frac{14}{5} .$

Normal at $(5, 6)$ : $y - 6 = \frac{14}{5} (x - 5) \Rightarrow 14 x - 5 y - 40 = 0.$

4. Length of Tangent and Power of a Point

The length of the tangent from an external point $(x_{1}, y_{1})$ to the circle

S \equiv x^{2} + y^{2} + 2 g x + 2 f y + c = 0

L = x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c = S_{1} .

Figure 4:

△ P T_{1} C

is right angled at

T_{1}

, so

P T_{1} = C P^{2} - r^{2}

. Substituting

C P^{2} = (x_{1} + g)^{2} + (y_{1} + f)^{2}

and

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

gives

L = S_{1}

Power of a point. The square of the length of the tangent from a point

P

to a circle is called the power of $P$ with respect to that circle. Power is

positive if $P$ lies outside the circle,
zero if $P$ lies on the circle,
negative if $P$ lies inside the circle.

It equals

S_{1}

(with the same sign) and is constant regardless of the chord/secant used to measure it.

Solved Example 6

Find the length of the tangent drawn from

(5, 1)

to the circle

x^{2} + y^{2} + 6 x - 4 y - 3 = 0

Solution:

$L = 5^{2} + 1^{2} + 6 (5) - 4 (1) - 3 = 25 + 1 + 30 - 4 - 3 = 49 = 7.$

5. Pair of Tangents from an External Point

From an external point $A (x_{1}, y_{1})$ we can draw exactly two tangents to a circle. Their combined equation (a second-degree "pair of lines") is

S S_{1} = T^{2}

where

$S \equiv x^{2} + y^{2} + 2 g x + 2 f y + c$ (the circle expression, with variables $x, y$ )
$S_{1} \equiv x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c$ (the same, evaluated at $(x_{1}, y_{1})$ ; a number)
$T \equiv x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c$ (the $T$ -form, linear in $x, y$ )

Solved Example 7

Find the equation of the pair of tangents drawn to the circle

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

from the point

(0, 1)

Solution:

$S \equiv x^{2} + y^{2} - 2 x + 4 y$ , $g = - 1, f = 2, c = 0$ .

$S_{1} = 0 + 1 - 0 + 4 = 5 > 0$ , so $(0, 1)$ lies outside.

$T \equiv 0 + y - (x + 0) + 2 (y + 1) + 0 = - x + 3 y + 2$ .

Pair of tangents $S S_{1} = T^{2}$ :

$5 (x^{2} + y^{2} - 2 x + 4 y) = (- x + 3 y + 2)^{2}$

Expand and simplify: $2 x^{2} - 2 y^{2} + 3 x y - 3 x + 4 y - 2 = 0.$

Factoring gives the separate tangents $2 x - y + 1 = 0$ and $x + 2 y - 2 = 0.$

6. Director Circle

The locus of points from which two perpendicular tangents can be drawn to a given circle is called the director circle. It is concentric with the given circle and has radius $2$ times the original.

Figure 5: The director circle is concentric with the given circle and has radius

r 2

. From any point

A

on it the two tangents to the inner circle meet at

9 0^{\circ}

Proof. Let the given circle have centre $C$ and radius $r$ . If from $A$ the two tangents are perpendicular, they touch the circle at $P$ and $Q$ so that $∠ P A Q = 9 0^{\circ}$ and $C P = C Q = r$ with $C P ⊥ A P$ , $C Q ⊥ A Q$ . Then $A P C Q$ is a square, and $A C = r csc 4 5^{\circ} = r 2$ .

Solved Example 8

Find the equation of the director circle of

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

Solution:

Given circle: centre $(2, - 1)$ , radius $2$ .

Director circle: same centre, radius $2 \cdot 2 = 2$ .

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

7. Chord of Contact

If two tangents from an external point $P (x_{1}, y_{1})$ touch the circle at $T_{1}$ and $T_{2}$ , the line $T_{1} T_{2}$ is called the chord of contact. Its equation is

T = 0, i.e. x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0.

Figure 6: The line joining the two points of contact

T_{1}

and

T_{2}

is the chord of contact of the tangents from

P (x_{1}, y_{1})

. Its equation is

T = 0

Useful formulas (let $R$ = radius, $L$ = length of tangent from $P$ ):

Chord of contact exists only if $P$ is not inside the circle.
Length of chord of contact $T_{1} T_{2} = \frac{2 L R}{R ^{2} + L ^{2}}$ .
Area of $△ P T_{1} T_{2} = \frac{R L ^{3}}{R ^{2} + L ^{2}}$ .
Tangent of the angle between the two tangents from $P$ $= \frac{2 R L}{L ^{2} - R ^{2}}$ .
Circle circumscribing $△ P T_{1} T_{2}$ : $(x - x_{1}) (x + g) + (y - y_{1}) (y + f) = 0$ (uses $C P$ as diameter).

Solved Example 9

Find the equation of the chord of contact of the tangents drawn from

(1, 2)

to the circle

x^{2} + y^{2} - 2 x + 4 y + 7 = 0

Solution:

$S_{1} = 1 + 4 - 2 + 8 + 7 = 18 > 0$ , so $(1, 2)$ lies outside. Chord of contact exists.

$T \equiv x (1) + y (2) - (x + 1) + 2 (y + 2) + 7 = x + 2 y - x - 1 + 2 y + 4 + 7 = 4 y + 10.$

Chord of contact: $4 y + 10 = 0$ , i.e. $2 y + 5 = 0.$

Solved Example 10

Tangents are drawn to the circle

x^{2} + y^{2} = 12

at the points where it is met by the circle

x^{2} + y^{2} - 5 x + 3 y - 2 = 0

; find the point of intersection of these tangents.

Solution:

Common chord of the two circles: $S_{1} - S_{2} = 0 \Rightarrow 5 x - 3 y - 10 = 0$ ...(i).

Suppose the tangents to the first circle at the two intersection points meet at $P (α, β)$ . Then the common chord is also the chord of contact of $P$ w.r.t. $x^{2} + y^{2} = 12$ , i.e.

$α x + β y - 12 = 0$ ...(ii).

Since (i) and (ii) are the same line, coefficients are proportional:

$\frac{α}{5} = \frac{β}{- 3} = \frac{- 12}{- 10} = \frac{6}{5} .$

So $α = 6, β = - \frac{18}{5}$ . Point of intersection: $(6, - \frac{18}{5}) .$

8. Pole and PolarJEE ADVANCED

Fix a point $P$ in the plane of a circle. Draw any secant through $P$ meeting the circle at $Q$ and $R$ . The tangents at $Q$ and $R$ meet at some point; as the secant rotates about $P$ , this meeting point traces a straight line. This line is called the polar of $P$ , and $P$ is called the pole of the line.

Equation of the polar of $P (x_{1}, y_{1})$ :

w.r.t. $x^{2} + y^{2} = a^{2}$ : $x x_{1} + y y_{1} = a^{2}$
w.r.t. general circle: $T = 0$ , same equation form as the tangent at a point.

If $P$ lies on the circle, its polar coincides with the tangent at $P$ . If $P$ lies outside the circle, its polar coincides with the chord of contact from $P$ .

Figure 7: The polar of

P

(purple) in the three possible positions of the pole. In every case its equation is the same,

T = 0

; only its position relative to the circle changes.

Pole of a given line

The pole of the line $A x + B y + C = 0$ with respect to $x^{2} + y^{2} = a^{2}$ is

(- \frac{A a ^{2}}{C}, - \frac{B a ^{2}}{C}) .

JEE ADVANCED Properties of pole and polar.

If the polar of $P$ passes through $Q$ , then the polar of $Q$ passes through $P$ (reciprocity).
Two points $P$ and $Q$ are called conjugate points if the polar of one passes through the other.
Two lines are called conjugate lines if the pole of one lies on the other.

Solved Example 11

Find the equation of the polar of

(2, - 1)

with respect to the circle

x^{2} + y^{2} - 3 x + 4 y - 8 = 0

Solution:

Apply $T = 0$ with $(x_{1}, y_{1}) = (2, - 1)$ , $g = - 3/2, f = 2, c = - 8$ :

$2 x + (- 1) y - \frac{3}{2} (x + 2) + 2 (y - 1) - 8 = 0$

$\Rightarrow 4 x - 2 y - 3 x - 6 + 4 y - 4 - 16 = 0 \Rightarrow x + 2 y - 26 = 0.$

Solved Example 12

Find the pole of the line

3 x + 5 y + 17 = 0

with respect to the circle

x^{2} + y^{2} + 4 x + 6 y + 9 = 0

Solution:

Let the pole be $P (α, β)$ . Polar of $P$ : $T = 0$ :

$α x + β y + 2 (x + α) + 3 (y + β) + 9 = 0$

$\Rightarrow (α + 2) x + (β + 3) y + (2 α + 3 β + 9) = 0.$

Compare with $3 x + 5 y + 17 = 0$ :

$\frac{α + 2}{3} = \frac{β + 3}{5} = \frac{2 α + 3 β + 9}{17} .$

From the first two: $5 (α + 2) = 3 (β + 3) \Rightarrow 5 α - 3 β = - 1$ .

From the first and third: $17 (α + 2) = 3 (2 α + 3 β + 9) \Rightarrow 11 α - 9 β = - 7.$

Solving: $α = 1, β = 2.$ Pole is $(1, 2) .$

9. Chord of a Circle with a Given Midpoint

The equation of the chord of $S \equiv x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ whose midpoint is $M (x_{1}, y_{1})$ is

T = S_{1},

i.e. $x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c .$

Figure 8: The chord with a given midpoint

M (x_{1}, y_{1})

is perpendicular to

C M

and is bisected at

M

. Its equation takes the compact form

T = S_{1}

Two facts about midpoint chords.

The shortest chord through a given interior point $M$ is the chord whose midpoint is $M$ (perpendicular to $C M$ ).
Equivalently, this is the chord whose distance from the centre is maximum among all chords through $M$ .

Chord joining two parametric points

The chord of $x^{2} + y^{2} = a^{2}$ joining the parametric points $P (α)$ and $Q (β)$ is

x cos \frac{α + β}{2} + y sin \frac{α + β}{2} = a cos \frac{α - β}{2} .

Setting $α = β$ recovers the parametric tangent $x cos α + y sin α = a$ .

Solved Example 13

Find the equation of the chord of

x^{2} + y^{2} + 6 x + 8 y - 11 = 0

whose midpoint is

(1, - 1)

Solution:

$S_{1} = 1 + 1 + 6 - 8 - 11 = - 11.$

$T \equiv x (1) + y (- 1) + 3 (x + 1) + 4 (y - 1) - 11 = x - y + 3 x + 3 + 4 y - 4 - 11 = 4 x + 3 y - 12.$

Chord ( $T = S_{1}$ ): $4 x + 3 y - 12 = - 11 \Rightarrow 4 x + 3 y - 1 = 0.$

Solved Example 14

Find the equation of the chord of

x^{2} + y^{2} = 15

which is bisected at

(3, 2)

Solution:

$S_{1} = 9 + 4 - 15 = - 2. T = 3 x + 2 y - 15.$

$T = S_{1}$ : $3 x + 2 y - 15 = - 2 \Rightarrow 3 x + 2 y - 13 = 0.$

Summary: Which Equation to Use When

Almost every question on this page is decided by one thing: what the point $(x_{1}, y_{1})$ is, and where it lies. Fix that first, then read off the equation.

What you are asked for	Equation	Condition on $S_{1}$
Tangent at a point on the circle	$T = 0$	$S_{1} = 0$
Chord of contact from an external point	$T = 0$	$S_{1} > 0$
Polar of a point	$T = 0$	any $S_{1}$ (point not the centre)
Chord bisected at a given point	$T = S_{1}$	$S_{1} < 0$ (point inside)
Pair of tangents from an external point	$S S_{1} = T^{2}$	$S_{1} > 0$
Length of tangent from a point	$L = S_{1}$	$S_{1} \geq 0$
Normal at a point on the circle	line through $(x_{1}, y_{1})$ and $(- g, - f)$	$S_{1} = 0$

One-line memory hook.

T = 0

is the "touching" family (tangent, chord of contact, polar).

T = S_{1}

is the "midpoint" case.

S S_{1} = T^{2}

is the only one that is second degree, because it represents two lines at once.

Common Mistakes to Avoid

Watch out

Using $T = 0$ instead of $T = S_{1}$ for the midpoint chord. $T = 0$ is the polar / tangent / chord-of-contact equation. The midpoint chord uses $T = S_{1}$ (with $S_{1} \neq = 0$ in general).
Forgetting to verify the point is external before writing $S S_{1} = T^{2}$ or the chord of contact. If $S_{1} \leq 0$ the "pair of tangents" and "chord of contact" don't have their usual geometric meaning.
Wrong slope for the normal. The normal at $(x_{1}, y_{1})$ passes through the centre $(- g, - f)$ , so its slope is $\frac{y _{1} + f}{x _{1} + g}$ - not $\frac{y _{1} - f}{x _{1} - g}$ (sign error is very common).
Using $y = m x + c$ tangent condition on a non-origin circle. The condition $c^{2} = a^{2} (1 + m^{2})$ is only for $x^{2} + y^{2} = a^{2}$ . For a general circle, use "perpendicular distance from centre = radius".
Confusing polar and chord of contact. Both have equation $T = 0$ , but "chord of contact" only makes sense when the point is external; "polar" is defined for every point (except the centre).
Squaring away signs in the length of tangent. $L = S_{1}$ needs $S_{1} \geq 0$ ; if you get $S_{1} < 0$ , the point is inside and there is no tangent.
Forgetting to compare all three ratios when finding a pole. Using only two of the three coefficient-ratios (from $T = 0$ vs the given line) can leave the pole undetermined; always use two independent pairs.

Frequently Asked Questions

What is the equation of a tangent to a circle at a given point?

For $x^{2} + y^{2} = a^{2}$ the tangent at $(x_{1}, y_{1})$ is $x x_{1} + y y_{1} = a^{2}$ . For the general circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , it is $x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c = 0$ , often written as $T = 0$ .

How do you find the length of a tangent from an external point to a circle?

Compute $L = S_{1}$ where $S_{1} = x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c$ evaluated at the external point $(x_{1}, y_{1})$ . $S_{1}$ must be positive; if it isn't, the point is inside or on the circle and no tangent exists.

What is the equation of the pair of tangents from an external point?

$S S_{1} = T^{2}$ , where $S$ is the circle expression, $S_{1}$ is its value at the external point (a number), and $T$ is $x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c$ . This gives a second-degree equation representing the two tangent lines together.

What is the chord of contact of a circle?

When two tangents are drawn from an external point $P (x_{1}, y_{1})$ to a circle, the line joining the two points of contact is the chord of contact. Its equation is $T = 0$ , the same as the polar of $P$ .

What is the director circle and what is its equation?

The locus of points from which two perpendicular tangents can be drawn to a circle is its director circle. It is concentric with the given circle and has radius $2$ times as large. For $x^{2} + y^{2} = a^{2}$ , the director circle is $x^{2} + y^{2} = 2 a^{2}$ .

What is the difference between a pole, polar and chord of contact?

The polar of a point $P$ with respect to a circle is the line whose pole is $P$ ; equation is $T = 0$ . If $P$ is external, the polar coincides with the chord of contact of the two tangents from $P$ . If $P$ is on the circle, the polar coincides with the tangent at $P$ .

How do you find the equation of a chord with a given midpoint?

Use $T = S_{1}$ , where $T$ is the tangent-form expression $x x_{1} + y y_{1} + g (x + x_{1}) + f (y + y_{1}) + c$ evaluated symbolically, and $S_{1}$ is the value of the circle expression at the midpoint $(x_{1}, y_{1})$ .

What is the length of a chord cut off from a circle by a line?

If the perpendicular distance from the centre to the line is $p$ and the radius is $r$ , the chord length is $2 r^{2} - p^{2}$ . This works because the perpendicular from the centre bisects the chord.

Why does the normal to a circle always pass through the centre?

The tangent at any point $P$ on a circle is perpendicular to the radius $C P$ . The normal at $P$ is the line perpendicular to the tangent through $P$ , which is exactly the line along the radius. Hence every normal to a circle passes through the centre.

Previous year questions on Circle: Chord, Tangent And Normal

21 questions from past papers, each with a step-by-step solution.

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