SSC CGL Maths · Circles

Tangents from an External Point

A tangent meets the radius at its point of contact at a right angle, so the radius, the tangent and the line from the outside point to the centre make a right triangle.

All Circles notes

Why this matters

Thirty-two questions in the bank, 27 EASY and 5 MODERATE. Half of them are one right triangle with the line to the centre as the hypotenuse. The rest use the two equal tangents from one point, the tangents from the corners of a triangle, and the alternate segment theorem, which carries all the MODERATE angle work.

Concept 1 of 4: Tangent at right angles to the radius, and its length

A tangent only grazes the circle. The radius to the point where it touches is the shortest way from the centre to that line, and the shortest way to a line is always at a right angle to it.

Definition

A tangent touches the circle at exactly one point, the point of contact. A line through two points of the circle is a secant, not a tangent.
The radius to the point of contact is perpendicular to the tangent. This holds for every circle, and for the diameter through that point too.
From an outside point P at distance $d$ from the centre, the tangent $t$ , the radius $r$ and $d$ form a right triangle with $d$ as the hypotenuse: $d^2 = r^2 + t^2$ .
Two of $d$ , $r$ , $t$ give the third. Find $r$ first when the question asks for the diameter or the area of the circle.
Fast route: the triples 3, 4, 5 and 5, 12, 13 and 8, 15, 17 and 7, 24, 25 settle most of these at sight. For an approximate answer, place the square between two perfect squares.

Length of a tangent

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

tthe length of the tangent from the outside point
dthe distance from the outside point to the centre
rthe radius

Worked example

A point P is 29 cm from the centre of a circle of radius 20 cm. Find the length of the tangent from P.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 19 Sep 2025, Shift 3 · Q69Easy

Example 1 · Circles · Tangents from an External Point

A point P is 13 cm away from the center of a circle. A tangent is drawn from P to the circle, and its length is 12 cm. What is the area of the circle?

Adding the squares when the distance is given

The line from the outside point to the centre is the hypotenuse. With that distance and the radius known, subtract the squares to get the tangent.

Answering with the radius when the diameter is asked

The right triangle gives the radius. A question about the diameter or the area of the circle needs one more step.

Drill 14 more on tangent at right angles to the radius, and its length

Concept 2 of 4: Two tangents from one outside point

The two tangents from P make two right triangles, OAP and OBP, that share the side OP and have equal radii. The triangles are congruent, so everything about one tangent is true of the other.

Definition

The two tangents from an outside point are equal: PA = PB.
OP bisects both $\angle APB$ and $\angle AOB$ .
In quadrilateral OAPB the angles at A and B are $90^\circ$ , so $\angle APB + \angle AOB = 180^\circ$ .
If $\angle APB = 90^\circ$ , OAPB is a square and $OP = r\sqrt{2}$ . If $\angle APB = 60^\circ$ , then $\angle APO = 30^\circ$ and $OP = 2r$ .
Area of OAPB is two right triangles: $2 \times \frac{1}{2} \times r \times t = r \times t$ .
The tangents at the two ends of a chord AB meet at T on the line through O and the chord's midpoint M. Triangles OMA and OAT are similar, so $TA = \frac{r \times AM}{OM}$ .

Two tangents from P

PA = PB \qquad \angle APB + \angle AOB = 180^\circ \qquad \text{Area}(OAPB) = r \times t

rthe radius
tthe length of each tangent

Worked example

Tangents PA and PB are drawn from an outside point P to a circle with centre O, and

\angle APB = 50^\circ

. Find

\angle AOB

and

\angle OPA

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 17 Sep 2025, Shift 3 · Q74Easy

Example 2 · Circles · Tangents from an External Point

When two tangents, PA and PB, are drawn to a circle from a point P outside the circle, and the angle between them is

90^\circ

,what is the distance from P to the center (O) of the circle, if the radius is 5 cm?

Setting the angle at the centre equal to the angle between the tangents

The two angles add to 180°, because the other two corners of OAPB are right angles. Tangents at 40° give 140° at the centre.

Using the whole angle between the tangents in triangle OAP

OP halves the angle between the tangents, so triangle OAP has half of it at P.

Drill 9 more on two tangents from one outside point

Concept 3 of 4: Tangent lengths from the corners of a triangle to its inscribed circle

A circle inside a triangle touching all three sides is touched by two tangents from each corner. Each corner's two tangents are equal, so the three sides are built from just three lengths.

Definition

Let the circle touch BC at D, CA at E and AB at F. Then $AE = AF$ , $BD = BF$ and $CD = CE$ .
Call these $x$ , $y$ and $z$ . The perimeter is $2(x + y + z)$ , so $x + y + z = s$ , the semi-perimeter.
The tangent from A is $s$ minus the side opposite A: $AF = s - BC$ . Likewise $BD = s - CA$ and $CE = s - AB$ .
Tangents PA and PB to a circle, and a third tangent touching the minor arc that cuts PA at R and PB at S: the perimeter of triangle PRS is $PA + PB = 2PA$ , because RS splits into two tangents that replace the pieces RA and SB.

Tangent from a corner

AF = s - a \qquad s = \frac{a + b + c}{2}

athe side BC, opposite the corner A
shalf the perimeter of the triangle

Worked example

A circle is inscribed in triangle ABC with AB = 12 cm, BC = 10 cm and CA = 8 cm. It touches AB at F. Find AF and BF.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 23 Sep 2025, Shift 3 · Q75Moderate

Example 3 · Circles · Tangents from an External Point

From an outside point P, two tangents, PA and PB, are drawn to a circle.A third tangent to the circle at a point Q intersects PA and PB at R and S respectively. If PA=10 cm, what is the perimeter of

\triangle PRS

Subtracting a neighbouring side from the semi-perimeter

The tangent from a corner is the semi-perimeter minus the side opposite that corner, the one side that does not touch it.

Adding the tangent lengths once for the perimeter

Every tangent length appears in two sides, so the perimeter is twice the sum of the three tangent lengths.

Drill 2 more on tangent lengths from the corners of a triangle to its inscribed circle

Concept 4 of 4: The alternate segment theorem

Slide a point along the circle towards C and the angle it sees chord CA at stays the same. When the point reaches C, the line to it has become the tangent, so the tangent makes that same angle with the chord.

Definition

Let a tangent touch the circle at C and let CA be a chord. The angle between the tangent and the chord equals the angle the chord subtends at any point B in the other segment (the arc on the far side of the chord from that angle).
The same angle is half the centre angle on the arc CA that lies inside the tangent-chord angle.
A quick check with the radius: $\angle OCA = 90^\circ -$ the tangent-chord angle, because the radius OC is perpendicular to the tangent.
In a figure with a tangent and a secant from one point, pair this theorem with the exterior angle of a triangle (an exterior angle equals the sum of the two opposite interior angles).

Tangent and chord

\angle(\text{tangent}, CA) = \angle CBA = \frac{1}{2}\angle COA

Ba point on the arc on the other side of the chord CA
Othe centre

Worked example

PT is a tangent to a circle with centre O at T, and TA is a chord with

\angle PTA = 48^\circ

, the angle opening towards the minor arc TA. B is a point on the major arc. Find

\angle TBA

and

\angle TOA

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 25 Sep 2024, Shift 3 · Q72Moderate

Example 4 · Circles · Tangents from an External Point

From a point T, a tangent TP at point P, is drawn to a circle with centre O. A secant TQR (point Q is near to point T) is drawn from the point T.

\Delta PQR

is inscribed into the circle by joining the points P, Q and R. Draw lines OQ and OR. If

\angle PTQ

27^\circ

and

\angle TPQ = 55^\circ

, what is the degree measure of

\angle ROQ

Using the angle in the same segment instead of the alternate one

The tangent-chord angle equals the angle the chord subtends on the far side of the chord. A point on the near side sees 180° minus it.

Doubling the tangent-chord angle to get the angle at the circumference

The tangent-chord angle already equals the angle at the circumference. Only the centre angle is twice it.

Drill 3 more on the alternate segment theorem

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 right angles to the radius, and its length
Length of a tangent
$t = \sqrt{d^2 - r^2}$
Two tangents from one outside point
Two tangents from P
$PA = PB \qquad \angle APB + \angle AOB = 180^\circ \qquad \text{Area}(OAPB) = r \times t$
Tangent lengths from the corners of a triangle to its inscribed circle
Tangent from a corner
$AF = s - a \qquad s = \frac{a + b + c}{2}$
The alternate segment theorem
Tangent and chord
$\angle(\text{tangent}, CA) = \angle CBA = \frac{1}{2}\angle COA$

Watch out for (8)

Adding the squares when the distance is given
→ Tangent at right angles to the radius, and its length
Answering with the radius when the diameter is asked
→ Tangent at right angles to the radius, and its length
Setting the angle at the centre equal to the angle between the tangents
→ Two tangents from one outside point
Using the whole angle between the tangents in triangle OAP
→ Two tangents from one outside point
Subtracting a neighbouring side from the semi-perimeter
→ Tangent lengths from the corners of a triangle to its inscribed circle
Adding the tangent lengths once for the perimeter
→ Tangent lengths from the corners of a triangle to its inscribed circle
Using the angle in the same segment instead of the alternate one
→ The alternate segment theorem
Doubling the tangent-chord angle to get the angle at the circumference
→ The alternate segment theorem

Test yourself on Circles

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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