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CDS Mathematics · Formula sheet

Circles formulas

14 formulas and 19 common traps for CDS Mathematics Circles, grouped by subtopic.

Full notes with worked examples

Chords and the Perpendicular from the Centre

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Half-chord, distance and radius

Chord and distance

r2=d2+(c2)2r^2 = d^2 + \left(\dfrac{c}{2}\right)^2

Arches, segment heights and equal chords

Radius from a segment

r=(c/2)2+h22hr = \dfrac{(c/2)^2 + h^2}{2h}

Common traps

Parallel chords have two answers

Unless the question says which side of the centre each chord lies on, both the difference and the sum of the distances are possible. The options often pair them, as in '2 cm or 14 cm'.

Half the chord, not the chord

The leg of the right triangle is HALF the chord. A 2424 cm chord in a circle of radius 1313 is 55 cm from the centre, not 242−132\sqrt{24^2 - 13^2}.

The centre is r − h from the chord

The height is measured from the chord to the arc, so the centre sits r−hr - h from the chord, not hh. Using hh makes the leg the wrong side of the triangle.

Radius or diameter?

Segment questions often ask for the DIAMETER. The equation gives rr; double it before matching options.

Angles in a Circle

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Angle at the centre and angles in the same segment

Inscribed angle

∠AOB=2 ∠ACB\angle AOB = 2\,\angle ACB

Cyclic quadrilaterals and the two segments

Cyclic quadrilateral

∠A+∠C=∠B+∠D=180∘\angle A + \angle C = \angle B + \angle D = 180^\circ

Common traps

An obtuse inscribed angle needs the reflex angle

Doubling an obtuse ∠ABC\angle ABC gives the REFLEX angle at the centre. Doubling 130∘130^\circ to 260∘260^\circ and stopping there answers the wrong angle; ∠AOC\angle AOC is 100∘100^\circ.

Two diameters meet at the centre

If two diameters meet at PP, then PP is the centre, and every segment from PP to the circle is a radius. That makes the triangles isosceles.

Segment, not sector

The textbook rule is about the angle in a SEGMENT. One paper printed 'sector'; a sector bigger than a semicircle has a reflex angle at the centre. Read the word before judging the statement.

The Circle Through Three Points, and Locus

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Circumcircle of a triangle

Circumradius

R=abc4ΔR = \dfrac{abc}{4\Delta}

Locus of a moving point

Sum of squares

PA2+PB2=2 PM2+AB22PA^2 + PB^2 = 2\,PM^2 + \dfrac{AB^2}{2}

Common traps

Equal distance from every vertex means circumcentre

A point 'at the same distance from each vertex' is the circumcentre, not the incentre. The circle it centres may be a different, smaller circle; the sides are then chords of the circumcircle.

Sum of squares is a circle, not a line

Equal DISTANCES from AA and BB give the perpendicular bisector. A constant SUM OF SQUARES of the distances gives a circle about the midpoint; the bisector is the distractor.

Tangents from an External Point

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Length of a tangent

Tangent length

PT=OP2−r2PT = \sqrt{OP^2 - r^2}

The alternate segment theorem

Tangent–chord angle

∠QPT=∠PRQ=12∠POQ\angle QPT = \angle PRQ = \tfrac12\angle POQ

A circle inside a quadrilateral

Tangential quadrilateral

AB+CD=BC+DAAB + CD = BC + DA

Common traps

Supplement, not double

The angle between the radii to the contact points is 180∘180^\circ minus the angle between the tangents, because the quadrilateral has two right angles. It is not twice that angle.

The ALTERNATE segment

The equal inscribed angle is on the far side of the chord from the tangent–chord angle. An angle in the near segment is its supplement.

Opposite sides, not adjacent ones

The rule pairs ABAB with CDCD and BCBC with DADA. Adding adjacent sides gives nothing.

Intersecting Chords and Power of a Point

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Chords, secants and the tangent from one point

Tangent and secant

PT2=PA⋅PBPT^2 = PA \cdot PB

Common traps

The whole secant, then subtract

In PT2=PA⋅PBPT^2 = PA \cdot PB, PBPB runs from PP to the FAR point. The chord asked for is PB−PAPB - PA; stopping at PBPB picks the distractor.

Given the product, the statements are not needed

If AP⋅PBAP \cdot PB is already given, CP⋅PDCP \cdot PD equals it by the theorem. In a data-sufficiency item, that means NEITHER statement is required.

Two Circles: Common Tangents and Common Chords

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Direct and transverse common tangents

Common tangents

direct=d2−(r1−r2)2,transverse=d2−(r1+r2)2\text{direct} = \sqrt{d^2 - (r_1 - r_2)^2}, \quad \text{transverse} = \sqrt{d^2 - (r_1 + r_2)^2}

The common chord of intersecting circles

Distance between centres

d=r12−h2+r22−h2d = \sqrt{r_1^2 - h^2} + \sqrt{r_2^2 - h^2}

Common traps

Difference for direct, sum for transverse

The direct tangent keeps both circles on one side, so the leg is the DIFFERENCE of the radii. The transverse tangent crosses between them and uses the SUM. Swapping them is the standard wrong option.

Both centres can be on one side

When the smaller circle's centre lies on the same side of the chord as the larger one's, dd is the DIFFERENCE of the two distances. Questions mean the usual picture, centres on opposite sides, unless they say otherwise.

Touching Circles

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Radii from contact and areas

Difference of the radii

(r1−r2)2=2(r12+r22)−(r1+r2)2(r_1 - r_2)^2 = 2(r_1^2 + r_2^2) - (r_1 + r_2)^2

Circles inside an angle

Two circles in an angle

r2r1=1−sin⁡θ1+sin⁡θ\dfrac{r_2}{r_1} = \dfrac{1 - \sin\theta}{1 + \sin\theta}

Common traps

Internal contact uses the difference

When one circle touches the other from inside, the centres are r1−r2r_1 - r_2 apart. Using the sum gives radii that do not fit the areas.

Diameters, not radii

Some items ask for the difference of the DIAMETERS, which is twice the difference of the radii.

Sine of the HALF-angle

The bisector splits the angle, so r=AOsin⁡θr = AO\sin\theta uses half the angle between the lines. With the full angle, a right angle would give r=AOr = AO.

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