Why is a tangent perpendicular to the radius at the point of contact?

This can be seen by shrinking a chord down to a single point: if a chord's central angle is y° and the angle it makes with a diameter at the fixed endpoint is x°, then 2x + y = 180°. As the second point slides to the first and y approaches 0, x approaches 90° — giving the tangent, perpendicular to the radius.

Are the two tangents drawn from an external point equal in length?

Yes. If PA and PB are tangents from an external point P to a circle centred at O, triangles OAP and OBP are congruent by RHS (OA=OB as radii, OP common, right angles at A and B), so PA = PB.

What is the relationship between the angle between two tangents and the angle between the radii?

If tangents from an external point touch the circle at A and B, then quadrilateral OACB (O = centre, C = external point) is cyclic, since the angles at A and B are both 90°. This means ∠AOB + ∠ACB = 180° — the angle between the tangents and the angle between the radii are always supplementary.

What is the alternate segment theorem?

The angle between a tangent and a chord drawn from the point of contact equals the angle subtended by that chord in the alternate segment (the region on the opposite side of the chord). Both angles equal half the chord's central angle.

What is the Pitot theorem for a tangent quadrilateral?

If a quadrilateral is formed by four tangents to a circle, the sums of its opposite sides are equal: a + c = b + d. This follows from each side being the sum of two tangent segments from adjacent vertices, and tangents from the same point being equal.

How is the radius of a triangle's incircle related to its area?

The incircle's radius r satisfies A = r × s, where A is the triangle's area and s is its semi-perimeter (half the total perimeter). So r = A/s. Combined with Heron's formula, this gives the incircle radius directly from the three side lengths.

Chapter 8 · Kerala SSLC Class 10 Maths

Tangents

Two ways to learn — a crisp formula reference, or a story that builds tangent theorems from a single shrinking chord, all the way to a circle inscribed inside a triangle.

Tangent ⊥ radiusEqual tangentsAlternate segment theoremPitot theoremIncircle

Also useful:Chapter 2 — Circles and Angles Chapter 10 — Circles And Lines Practice Problems — Ch 8 (solved questions)1-on-1 Kerala SSLC Maths Tuition

Kerala SSLCClass 10MathematicsChapter 8⏱ 13 min crisp · 19 min story

Six scenes from a shrinking chord to a circle inscribed in a triangle.

How this works

Six scenes build tangents from a single shrinking chord all the way to a circle inscribed inside a triangle. Read the story, then tap "The Maths" for the formal result.

The Shrinking Chord

Draw a circle and mark two points on it: A, fixed, and X, which we will slide closer and closer to A. Extend the diameter through A into a line, and draw the chord AX.

As X slides toward A, the chord shrinks — and the angle it makes with the diameter at A keeps changing. What does that angle approach as X reaches A?

Setting up the angles

Let the angle between the diameter and the chord AX at A be x°, and the central angle of the arc AX be y°.

OA and OX are both radii, so triangle OAX is isosceles — the angles opposite these equal sides are equal, meaning the third angle of the triangle is also x°.

Angles of triangle OAX: x° + x° + y° = 180°

2x + y = 180

As X slides closer to A, the chord shrinks and its central angle y gets closer to zero.

x = (180 − y) / 2 = 90 − y/2

As y → 0, x → 90°

So as the chord shrinks all the way down to a single touching line — the tangent — the angle x becomes exactly 90°.

Two Ropes from a Peg

A peg is fixed outside a circular fence. Two ropes are stretched taut from the peg until each just grazes the fence at a single point — these are the two tangents from the peg.

Claim

Both ropes are exactly the same length, no matter where the peg is placed.

Here's why. In triangle OAP and triangle OBP (O = centre, P = peg, A and B = points of contact):

OA = OB (radii of the same circle)

OP = OP (common side)

∠OAP = ∠OBP = 90° (tangent ⊥ radius, Scene 1)

∴ △OAP ≅ △OBP (RHS congruence)

∴ PA = PB

And since OP is common to both congruent triangles, it also bisects both ∠AOB and ∠APB.

One more curiosity: if the circle's radius is taken as exactly 1 unit, the length of a tangent segment from a point at angle x on the circle works out to be exactly tan x — which is exactly where trigonometry's "tan" gets its name (from the Latin tangere, "to touch").

Since OA ⊥ AP, if you know the radius and the distance from the peg to the centre, Pythagoras gives the tangent length directly.

OP² = OA² + AP²

AP = √(OP² − OA²)

The Angle That Splits in Two

Tangents are drawn at points A and B on a circle centred at O, meeting at an external point C. Look at the quadrilateral OACB.

The angles at A and B (between each tangent and its radius) are both 90°, so they already add up to 180° by themselves.

The consequence

A quadrilateral where one PAIR of opposite angles sums to 180° is a cyclic quadrilateral — and in ANY cyclic quadrilateral, both pairs of opposite angles sum to 180°.

∠OAC + ∠OBC = 90° + 90° = 180° → OACB is cyclic

So the OTHER pair also sums to 180°:

∠AOB + ∠ACB = 180°

So the angle between the two tangents, at the external point, and the central angle between the two radii, always add up to 180°.

A nice special case: if three tangents are drawn to a circle so that the radii to their points of contact are each 120° apart, the tangent lines form an EQUILATERAL triangle wrapped around the circle — because each angle between adjacent tangents is 180° − 120° = 60°.

The Sliding Secant

Draw a tangent at point T, and a chord TC from the same point. What is the relationship between the angle the chord makes with the tangent, and the arc it cuts off?

Let the central angle of chord TC be x°, and let y° be the angle the (lower) tangent makes with the chord at T.

Since the tangent is perpendicular to the radius OT, the angle between the radius OT and the chord TC is (90 − y)°. Triangle OTC is isosceles (OT = OC, both radii), so the angle at C is also (90 − y)°.

Angles of triangle OTC: (90−y) + (90−y) + x = 180

180 − 2y + x = 180

x = 2y → y = x/2

So the tangent-chord angle is always exactly HALF the chord's central angle.

Connecting to Circles and Angles

From the earlier chapter, any inscribed angle in the alternate segment is ALSO half the central angle of the same chord.

So both the tangent-chord angle and the inscribed angle in the alternate segment equal x/2 — meaning they must equal each other.

This theorem is powerful enough to draw a tangent at a point WITHOUT knowing where the circle's centre is: draw any chord through the point, measure the angle it makes with a point elsewhere on the circle, and reproduce that same angle on the other side of the chord.

The Kite of Four Tangents

Four tangents are drawn to a circle at four different points, forming a quadrilateral around the circle. Its sides are a, b, c, d, in order.

From each vertex, two tangent segments reach the circle, and — as in Scene 2 — a pair of tangents from the same external point are always equal. Label these tangent-segment lengths p, q, r, s around the quadrilateral.

a = p + q

b = q + r

c = r + s

d = s + p

Add the FIRST and THIRD sides together, and separately the second and fourth:

a + c = (p+q) + (r+s) = p + q + r + s

b + d = (q+r) + (s+p) = p + q + r + s

Both sums come out exactly the same.

Fitting a Circle Inside a Triangle

Reverse question: instead of drawing tangents to a fixed circle, can we draw a circle that touches two given lines meeting at a point?

Any point equidistant (measuring perpendicular distance) from both lines works as a centre — and the set of all such points is exactly the ANGLE BISECTOR between the two lines.

Stretch this to a full triangle: to touch all three sides, a circle's centre must be equidistant from all three sides at once — meaning it must lie on all THREE angle bisectors.

The key question

Do the three angle bisectors of a triangle always meet at one common point?

Yes — always. That common point is called the incentre, and the circle centred there, touching all three sides, is the incircle.

Now join the incentre to all three vertices, splitting the triangle into three smaller triangles. Each smaller triangle has one side of the original triangle as its base, and the incircle's radius r as its height.

Areas of the three smaller triangles: (1/2)ar, (1/2)br, (1/2)cr

Total area A = (1/2)r(a + b + c)

A = r × [(a+b+c)/2] = r × s (s = semi-perimeter)

So the incircle's radius is simply the triangle's area divided by its semi-perimeter.

The Big Picture

Every idea in this chapter grows out of one right angle. Master these seven moves and Chapter 8 is done.

Where does a tangent meet a radius?

At exactly 90° (OT ⊥ PT)

Two tangents from one external point?

PA = PB, and PT = √(OP² − r²)

Angle between two tangents + central angle?

sum = 180° (via cyclic OACB)

Angle between a tangent and a chord?

= angle in the alternate segment

Sides of a tangent quadrilateral?

a + c = b + d (Pitot theorem)

Circle touching all three sides of a triangle?

centred at the incentre

Incircle radius from area?

r = Area / s (s = semi-perimeter)

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