SSC CGL Maths · Circles

Intersecting Chords and Power of a Point

Every line through a fixed point P that meets a circle cuts it into two pieces whose product is the same, whether P is inside the circle, outside it, or at the end of a tangent.

All Circles notes

Why this matters

Eighteen questions in the bank, 11 EASY and 7 MODERATE. Each is one product equation. The marks are lost in choosing the pieces: inside the circle you multiply the two parts of a chord, outside you multiply the near length by the whole length from P.

Concept 1 of 3: Two chords crossing inside a circle

When chords AB and CD cross at P, triangles PAC and PDB have equal angles (vertically opposite angles at P, and angles in the same segment), so their sides are in proportion. That proportion says the two products of pieces are equal.

Definition

Chords AB and CD cross at P inside the circle. Then $PA \times PB = PC \times PD$ .
All four lengths are measured from P, the crossing point, to the circle.
Given a whole chord, subtract to get the second piece first. After finding the missing piece, add to get a whole chord if that is asked.
Pieces given as a ratio: write them as $2k$ and $3k$ , use the product rule, and the ratio of whole chords follows.

Intersecting chords

PA \times PB = PC \times PD

Pthe point where the two chords cross

Worked example

Chords AB and CD of a circle cross at P. AP = 15 cm, PB = 4 cm and CP = 6 cm. Find PD and CD.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 17 Sep 2025, Shift 2 · Q72Easy

Example 1 · Circles · Intersecting Chords and Power of a Point

Two chords AB and CD intersect at a point P inside a circle. If AP = 6 cm, PB = 8 cm, and CP = 4 cm, what is the length of PD?

Multiplying a piece by the whole chord when the chords cross inside

Inside the circle both factors are pieces measured from the crossing point. The whole chord enters only when you add the pieces at the end.

Drill 4 more on two chords crossing inside a circle

Concept 2 of 3: Secants and a tangent from an outside point

From a point P outside, a line through the circle meets it twice: first at a near point A, then at a far point B. The product of PA and PB is the same for every such line, and for a tangent the two points become one, so the product is the tangent squared.

Definition

Secants PAB and PCD from an outside point P: $PA \times PB = PC \times PD$ , where each factor runs from P: to the near point and to the far point.
The far length is the near length plus the chord: $PB = PA + AB$ .
With a tangent PT from the same point: $PT^2 = PA \times PB$ .
If the near piece $x$ is unknown and the chord $c$ is given, the rule gives $x(x + c) = t^2$ . Factorise and keep the positive root.
If the radius is given too, check the chord is not longer than the diameter. A chord longer than the diameter means the printed data cannot exist.

Secants and a tangent from P

PA \times PB = PC \times PD \qquad PT^2 = PA \times PB

PAP to the near point of the secant
PBP to the far point, that is PA plus the chord AB
PTthe tangent from P

Worked example

From an outside point P, a secant meets a circle at A and B, with A nearer P. PA = 5 cm and AB = 15 cm. Find the length of the tangent from P.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 19 Sep 2025, Shift 3 · Q74Moderate

Example 2 · Circles · Intersecting Chords and Power of a Point

From a point P outside a circle, a tangent PA and a secant PBC are drawn. If PA=6 cm and BC=5 cm, what is the length of the segment PB?

Using the chord instead of the whole secant

Outside the circle the second factor is the full distance from P to the far point, the near length plus the chord, never the chord alone.

Squaring the near length instead of the tangent

The tangent squared equals near length times far length. The near length is not a tangent and is never squared on its own.

Drill 10 more on secants and a tangent from an outside point

Concept 3 of 3: Power of a point from its distance to the centre

The product

PA \times PB

is the same for every line through P, so take the line through the centre. It meets the circle at distances

d - r

and

d + r

from P, and their product is

d^2 - r^2

, which is also the tangent squared.

Definition

P outside the circle, at distance $d$ from the centre: $PA \times PB = d^2 - r^2$ , the square of the tangent from P.
P inside: $PA \times PB = r^2 - d^2$ for every chord through P.
So a secant's two lengths and the distance to the centre give the radius at once, without drawing the tangent.

Power of a point

PA \times PB = d^2 - r^2 \text{ (P outside)} \qquad PA \times PB = r^2 - d^2 \text{ (P inside)}

dthe distance from P to the centre
rthe radius

Worked example

A point P is 13 cm from the centre of a circle of radius 5 cm. A secant from P meets the circle at A and B with PA = 8 cm. Find PB and AB.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 19 Sep 2024, Shift 1 · Q65Moderate

Example 3 · Circles · Intersecting Chords and Power of a Point

R is a point outside a circle and is 18 cm away from its centre. A secant drawn from the point R intersects the circle at points A and B in such a way that RA = 7 cm and AB = 5 cm. The radius of the circle (in cm) is:

Using $r^2 - d^2$ for an outside point

For a point outside the circle d is larger than r, so the product is

d^2 - r^2

. The order flips only for a point inside.

Drill 1 more on power of a point from its distance to the centre

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 (3)

Two chords crossing inside a circle
Intersecting chords
$PA \times PB = PC \times PD$
Secants and a tangent from an outside point
Secants and a tangent from P
$PA \times PB = PC \times PD \qquad PT^2 = PA \times PB$
Power of a point from its distance to the centre
Power of a point
$PA \times PB = d^2 - r^2 \text{ (P outside)} \qquad PA \times PB = r^2 - d^2 \text{ (P inside)}$

Watch out for (4)

Multiplying a piece by the whole chord when the chords cross inside
→ Two chords crossing inside a circle
Using the chord instead of the whole secant
→ Secants and a tangent from an outside point
Squaring the near length instead of the tangent
→ Secants and a tangent from an outside point
Using $r^2 - d^2$ for an outside point
→ Power of a point from its distance to the centre

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.

Take the Circles chapter test Drill all 18 questions