MHT-CET Maths · Circle

Tangents — At a Point, With a Given Slope, From an External Point and Their Loci

Tangent at (x₁, y₁): xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0; with slope m to x² + y² = a²: y = mx ± a√(1 + m²); from an external point the tangent length is √S₁ and the two tangents with the two radii make a kite.

All Circle notes MHT-CET Maths strategy

Why this matters

21 PYQs at 38% HARD — the chapter's second-largest page and the one with the most HARD questions. The tangent at the far end of a diameter, the parametric tangent, tangents of a given slope, a parabola's tangent that also touches a circle (twice), the kite PAOB area (three times), a tangent-length locus, the 60°-tangents locus, and the classical PQ · RS = (2r)² result. Half the marks are the kite: tangent length √S₁ times radius is the area, and sin of the half-angle is r over the distance.

Concept 1 of 4: Tangent at a Point on the Circle: T = 0

Replace

x^2 \to xx_1

y^2 \to yy_1

2x \to x + x_1

2y \to y + y_1

in the circle's equation — the result is the tangent at

(x_1, y_1)

. Equivalently, the tangent is perpendicular to the radius

CP

. At a parametric point on

x^2 + y^2 = a^2

it is

x\cos\theta + y\sin\theta = a

Definition

$x^2 + y^2 - 6x - 5y - 1 = 0$ , one end of a diameter $(-1, 3)$ : centre $\left(3, \tfrac52\right)$ , other end $(7, 2)$ ; tangent there: $7x + 2y - 3(x + 7) - \tfrac52(y + 2) - 1 = 0 \Rightarrow 8x - y - 54 = 0$ .
$x = 5\cos\theta$ , $y = 5\sin\theta$ at $\theta = \tfrac{\pi}{3}$ : $\dfrac{x}{2} + \dfrac{\sqrt3 y}{2} = 5 \Rightarrow x + \sqrt3 y = 10$ .
Tangent and normal at $(\sqrt3, 1)$ on $x^2 + y^2 = 4$ : tangent $\sqrt3 x + y = 4$ meets the $X$ -axis at $\left(\tfrac{4}{\sqrt3}, 0\right)$ ; the normal is through the origin; triangle area $\tfrac12 \cdot \tfrac{4}{\sqrt3} \cdot 1 = \tfrac{2}{\sqrt3}$ .
The normal at any point passes through the centre — a one-line fact that kills half the normal stems.

Tangent at a point

xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0 \qquad x^2 + y^2 = a^2:\ x\cos\theta + y\sin\theta = a

Worked example

Find the tangent to

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

(5, 1)

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q135Hard

Example 1 · Circle · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci

One end of the diameter of the circle

x^2+y^2-6x-5y-1=0

(-1,3)

, then the equation of the tangent at the other end of the diameter is

Forgetting to halve the linear coefficients in T

-6x

becomes

-3(x + x_1)

, not

-6(x + x_1)

. Doubling gives

8x - 2y - 52 = 0

-style options that are exactly the planted distractors.

Drill 2 more on tangent at a point on the circle: t = 0

Concept 2 of 4: Tangent of a Given Slope, and Whether a Line Touches: Distance From the Centre = Radius

y = mx + c

touches

x^2 + y^2 = a^2

iff

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

— which is just 'distance from the origin equals

a

'. For any circle, a line is tangent iff the centre's distance from it equals

r

; that one test also settles a tangent to another curve touching the circle.

Definition

Perpendicular to $5x + y = 2$ (so slope $\tfrac15$ ) and tangent to $x^2 + y^2 = 36$ : $y = \dfrac{x}{5} \pm 6\sqrt{1 + \tfrac{1}{25}} \Rightarrow x - 5y \pm 6\sqrt{26} = 0$ .
Tangent to $x^2 = y - 6$ at $(1, 7)$ is $2x - y + 5 = 0$ ; it touches $x^2 + y^2 + 16x + 12y + C = 0$ (centre $(-8, -6)$ , $r^2 = 100 - C$ ) iff $\dfrac{|-16 + 6 + 5|}{\sqrt5} = \sqrt{100 - C} \Rightarrow 5 = 100 - C \Rightarrow C = 95$ .
The tangent $x - 2y = 5$ to $x^2 + y^2 = 5$ at $(1, -2)$ also touches $x^2 + y^2 - 8x + 6y + 20 = 0$ (centre $(4, -3)$ , $r = \sqrt5$ ); the contact point is the foot of the perpendicular from $(4, -3)$ : $(3, -1)$ .
Two tangents of each slope: the $\pm$ is the two sides of the circle.

Tangency condition

y = mx \pm a\sqrt{1 + m^2} \qquad \text{general: } \frac{|ah + bk + c|}{\sqrt{a^2 + b^2}} = r

Worked example

Find the tangents to

x^2 + y^2 = 9

parallel to

3x + 4y = 1

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift II · Q116Moderate

Example 2 · Circle · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci

The equations of the tangents to the circle

x^{2}+y^{2}= 36

which are perpendicular to the line

5x + y = 2

, are

Using the slope of the given line instead of the perpendicular one

'Perpendicular to

5x + y = 2

' means slope

\dfrac15

, giving

x - 5y \pm 6\sqrt{26} = 0

. Options (C) and (D) use slope

-5

5

Drill 2 more on tangent of a given slope, and whether a line touches: distance from the centre = radius

Concept 3 of 4: Tangents From an External Point: Length √S₁, the Kite, the Angle Between Them

From

P

outside the circle, both tangent lengths equal

\sqrt{S_1}

, where

S_1

is the circle's expression evaluated at

P

— Pythagoras on

CP

r

and the tangent. The quadrilateral

PAOB

(two tangents, two radii) has area

r\sqrt{S_1}

, and half the angle between the tangents has

\sin

equal to

\dfrac{r}{CP}

Definition

$P(1, 7)$ , $x^2 + y^2 = 25$ : $\sqrt{S_1} = \sqrt{1 + 49 - 25} = 5$ ; area $PQOR = 2 \cdot \tfrac12 \cdot 5 \cdot 5 = 25$ .
$P(-4, 0)$ , $x^2 + y^2 = 4$ : $PA = \sqrt{16 - 4} = 2\sqrt3$ ; area $PAOB = 2 \cdot 2\sqrt3 \cdot \tfrac12 \cdot 2 = 4\sqrt3$ .
$P(-4, -5)$ , $x^2 + y^2 + 6x - 4y - 12 = 0$ (centre $(-3, 2)$ , $r = 5$ ): tangent length $5 = r$ , so the kite is a square of area $25$ and the sector inside it is a quarter circle; the area between the tangents and the circle is $25 - \tfrac{25\pi}{4} = 25\left(\tfrac{4 - \pi}{4}\right)$ .
Angle $60^\circ$ between the tangents to $x^2 + y^2 = 16$ : $\sin 30^\circ = \dfrac{4}{OP} \Rightarrow OP = 8$ ; locus of $P$ : $x^2 + y^2 = 64$ . (Director circle, $90^\circ$ : $x^2 + y^2 = 2a^2$ .)

External point

L = \sqrt{S_1} = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c},\qquad [PAOB] = rL,\qquad \sin\frac{\theta}{2} = \frac{r}{CP}

Worked example

Find the length of the tangent from

(6, 8)

x^2 + y^2 - 2x - 4y - 20 = 0

, and the area of the kite formed with the two radii.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q120Moderate

Example 3 · Circle · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci

Two tangents to the circle

x^2+y^2=4

at the points A and B meet at the point

P(-4,0)

. Then the area of the quadrilateral

PAOB

, O being the origin, is

Halving the kite

PAOB

is TWO right triangles, so its area is

rL

, not

\tfrac12 rL

2\sqrt3

is option (A) on the

(-4, 0)

stem; the answer is

4\sqrt3

Drill 4 more on tangents from an external point: length √s₁, the kite, the angle between them

Concept 4 of 4: Loci From Tangent Lengths, and PQ · RS = (2r)²

A ratio of tangent lengths to two circles is a ratio of

\sqrt{S_1}

values; square it and the locus is a circle (the

x^2 + y^2

terms do not cancel unless the ratio is

1

, which gives the radical axis, a line). For tangents at the ends of a diameter, the crossed lines meeting on the circle give

PQ \cdot RS = (2r)^2

Definition

Ratio $2 : 3$ to $x^2 + y^2 + 4x + 3 = 0$ and $x^2 + y^2 - 6x + 5 = 0$ : $9(x^2 + y^2 + 4x + 3) = 4(x^2 + y^2 - 6x + 5) \Rightarrow 5x^2 + 5y^2 + 60x + 7 = 0$ .
Ratio $1 : 1$ : $S_1 = S_2$ is the radical axis, a straight line.
$PQ$ and $RS$ tangents at the ends of diameter $PR$ , $PS$ and $RQ$ meeting at $X$ on the circle: $\angle PXR = 90^\circ$ , and the similar right triangles $PQR$ -type give $PQ \cdot RS = PR^2 = 4r^2$ , so $2r = \sqrt{PQ \cdot RS}$ .
Squaring a ratio of lengths is safe because both lengths are positive.

Tangent-length locus

\frac{\sqrt{S_1}}{\sqrt{S_2}} = \frac{m}{n} \iff n^2 S_1 = m^2 S_2 \qquad PQ \cdot RS = (2r)^2

Worked example

Find the locus of a point whose tangent lengths to

x^2 + y^2 = 4

and

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

are equal.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q108Hard

Example 4 · Circle · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci

The locus of a point which moves such that the ratio of the length of the tangents to the circles

x^2+y^2+4x+3=0

and

x^2+y^2-6x+5=0

2:3

, is

Sign of the x term after cross-multiplying

9(4x) - 4(-6x) = 36x + 24x = +60x

. The bank once carried

-60x

as its key; the sign is positive.

Drill 1 more on loci from tangent lengths, and pq · rs = (2r)²

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)

Tangent at a Point on the Circle: T = 0
Tangent at a point
$xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0 \qquad x^2 + y^2 = a^2:\ x\cos\theta + y\sin\theta = a$
Tangent of a Given Slope, and Whether a Line Touches: Distance From the Centre = Radius
Tangency condition
$y = mx \pm a\sqrt{1 + m^2} \qquad \text{general: } \frac{|ah + bk + c|}{\sqrt{a^2 + b^2}} = r$
Tangents From an External Point: Length √S₁, the Kite, the Angle Between Them
External point
$L = \sqrt{S_1} = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c},\qquad [PAOB] = rL,\qquad \sin\frac{\theta}{2} = \frac{r}{CP}$
Loci From Tangent Lengths, and PQ · RS = (2r)²
Tangent-length locus
$\frac{\sqrt{S_1}}{\sqrt{S_2}} = \frac{m}{n} \iff n^2 S_1 = m^2 S_2 \qquad PQ \cdot RS = (2r)^2$

Watch out for (4)

Forgetting to halve the linear coefficients in T
→ Tangent at a Point on the Circle: T = 0
Using the slope of the given line instead of the perpendicular one
→ Tangent of a Given Slope, and Whether a Line Touches: Distance From the Centre = Radius
Halving the kite
→ Tangents From an External Point: Length √S₁, the Kite, the Angle Between Them
Sign of the x term after cross-multiplying
→ Loci From Tangent Lengths, and PQ · RS = (2r)²

Test yourself on Circle

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Related notes

Straight Line — the foot of the perpendicular gives the point of contact →