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NDA Mathematics · Formula sheet

Circles formulas

20 formulas and 20 common traps for NDA Mathematics Circles, grouped by subtopic.

Full notes with worked examples

Circle Equation — Centre, Radius & Properties

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What a Circle Equation Is

Standard form

(x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2
  • (h,k)(h,k)centre
  • rrradius (diameter = 2r)

General Form — Centre and Radius by Completing the Square

General form

x2+y2+2gx+2fy+c=0  ⇒  centre (−g,−f),    r=g2+f2−cx^2+y^2+2gx+2fy+c=0 \;\Rightarrow\; \text{centre }(-g,-f),\;\; r=\sqrt{g^2+f^2-c}

Diameter Form — Circle From Two Endpoints

Diameter form

(x−x1)(x−x2)+(y−y1)(y−y2)=0(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0

Intercepts a Circle Cuts on the Axes

Axis intercept lengths

x-axis: 2g2−cy-axis: 2f2−c\text{x-axis: }2\sqrt{g^2-c}\qquad \text{y-axis: }2\sqrt{f^2-c}

Perpendicular From the Centre Bisects a Chord

Chord length from centre distance

chord=2r2−d2(d=distance from centre to the chord)\text{chord} = 2\sqrt{r^2 - d^2}\quad(d=\text{distance from centre to the chord})

Circles That Touch the Axes

Tangency condition

∣ah+bk+c∣a2+b2=r\frac{|ah+bk+c|}{\sqrt{a^2+b^2}} = r
  • (h,k)(h,k)centre
  • rrradius

Two Circles — Intersecting, Touching, Separate

Two distinct intersections

∣r1−r2∣<d<r1+r2|r_1 - r_2| < d < r_1 + r_2

Circle Through the Origin With Given Axis Intercepts

Circle through origin, intercepts a, b

x2+y2−ax−by=0,centre (a2,b2)x^2+y^2-ax-by=0,\qquad \text{centre }\left(\tfrac a2,\tfrac b2\right)

Common traps

Divide by the leading coefficient BEFORE reading g, f, c

A 4x2+4y2+…4x^2+4y^2+\ldots circle is the single most common NDA trap here. The centre is NOT (−g,−f)(-g,-f) of the un-divided equation — you must first make the x2x^2 coefficient 11. Skipping this scales the centre and radius by the wrong factor.

Centre is MINUS g and MINUS f

From x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0 the centre is (−g,−f)(-g,-f). Many slips come from reading the centre as (g,f)(g,f) or as (2g,2f)(2g,2f) — it is half the coefficient, negated.

The x-factors and y-factors are separate

In (x−x1)(x−x2)+(y−y1)(y−y2)=0(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0 the endpoints are (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2) — you pair the FIRST x-factor with the FIRST y-factor. A common error mixes them, e.g. reading endpoints as (x1,y2)(x_1,y_2), giving the wrong diameter.

Intercept is the GAP between roots, not a single root

After zeroing a variable you get two roots — the intercept length is ∣y1−y2∣|y_1-y_2| (or ∣x1−x2∣|x_1-x_2|), the distance between them. Reporting just one root, or their sum, is the standard slip. If the quadratic has no real roots, the circle simply doesn't meet that axis.

Use the NEGATIVE-reciprocal slope for the perpendicular

If the chord's line has slope mm, the line from the centre is perpendicular with slope −1/m-1/m — not mm, not 1/m1/m. Getting the sign or the reciprocal wrong lands you at the wrong point on the chord (the sign-of-slope slip is exactly what trips this PYQ).

Touching an axis is |coordinate| = r, not coordinate = r

A circle touching both axes can sit in any quadrant: centre (±r,±r)(\pm r,\pm r). The PYQ usually pins it to the first quadrant, giving (r,r)(r,r) — but read the quadrant condition. And touching a line means distance =r=r (tangent), which is stricter than merely crossing it.

Both inequalities matter — it's a band, not a single bound

"Intersect at two points" is the strict double inequality ∣r1−r2∣<d<r1+r2|r_1-r_2|<d<r_1+r_2. Using only d<r1+r2d<r_1+r_2 lets one circle sit entirely inside the other (which has NO intersection). Always check the lower bound too.

Through the origin forces the constant term to vanish

Substituting (0,0)(0,0) into x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0 gives c=0c=0 — a circle through the origin has no constant term. Forgetting this adds a spurious unknown and the system stops being solvable from the three points.

Circles Through Given Points & Concyclicity

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Building a Circle From Diameter Endpoints

Circle on a diameter

(x−x1)(x−x2)+(y−y1)(y−y2)=0(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0

Circle Through Three Points — the General-Equation System

Unknown-coefficient circle

x2+y2+Dx+Ey+F=0,centre (−D2,−E2)x^2+y^2+Dx+Ey+F=0,\quad \text{centre }\left(-\tfrac D2,-\tfrac E2\right)

Extracting Centre and Radius From Three Points

Radius from centre and a point

r2=(x0−h)2+(y0−k)2r^2 = (x_0-h)^2 + (y_0-k)^2
  • (h,k)(h,k)centre
  • (x0,y0)(x_0,y_0)any point on the circle

Centre on a Given Line — Perpendicular-Bisector Method

Equidistance condition

(h−x1)2+(k−y1)2=(h−x2)2+(k−y2)2(h-x_1)^2+(k-y_1)^2 = (h-x_2)^2+(k-y_2)^2

Concyclicity — Does a Fourth Point Lie on the Circle?

Concyclicity

(x4,y4) on x2+y2+Dx+Ey+F=0  ⟺  x42+y42+Dx4+Ey4+F=0(x_4,y_4)\text{ on }x^2+y^2+Dx+Ey+F=0 \iff x_4^2+y_4^2+Dx_4+Ey_4+F=0

Family of Circles Through a Chord (the S + λL Trick)

Family through a chord

S+λL=0S + \lambda L = 0

Circumcentre of a Right Triangle — Midpoint of the Hypotenuse

Right-triangle circumcentre

circumcentre=midpoint of hypotenuse,R=12(hypotenuse)\text{circumcentre} = \text{midpoint of hypotenuse},\quad R=\tfrac12(\text{hypotenuse})

Common traps

Clear fractions, then match the option's scale

The system often gives fractional D,E,FD,E,F. The answer options may be scaled up (e.g. 4x2+4y2+…4x^2+4y^2+\ldots) to clear them — multiply through to match, but remember the circle is the same. Don't read g=Dg=D; the general form uses 2g=D2g=D, so centre is −D/2-D/2 not −D-D.

Compute r² and compare squares — skip the root

If the question asks only whether rr beats a bound, compare r2r^2 with (bound)². Forcing a messy square root invites arithmetic slips. And always take the radius from a KNOWN point on the circle, not from a half-remembered formula.

Two through-points give ONE equation, not two

Equating the distances to the two given points yields a single line (the perpendicular bisector), so you still need the centre-on-a-line constraint to fix the point. With only the two points you'd have a whole family of circles — the extra line is what makes the answer unique.

An unknown coordinate gives TWO values — keep both

Substituting (0,k)(0,k) yields a quadratic, so there are usually two valid kk (one may coincide with a given point). The NDA answer often lists BOTH; discarding one because it 'looks like' an existing point loses a mark.

"Chord as diameter" = the new centre sits on the chord line

The condition that makes λ\lambda solvable is geometric: the chord is a diameter of the new circle exactly when the new centre lies ON the chord line LL. Trying to force the new circle to pass through a chord endpoint instead leaves λ\lambda undetermined.

Only works when there IS a right angle

The midpoint-of-hypotenuse shortcut needs a right-angled triangle. Confirm two sides are perpendicular first (perpendicular lines, or slopes multiplying to −1-1). For a general triangle you must intersect two perpendicular bisectors instead.

Inscribed Geometry, Tangents & Segments

Learn this subtopic in the notes

Inscribed Angle and the Angle in a Semicircle

Inscribed angle

∠BAC=12 ∠BOC\angle BAC = \tfrac12\,\angle BOC
  • OOcentre
  • AApoint on the circle

Points Where a Circle Touches the Axes

Contact points and PQ

P=(k,0),    Q=(0,k),    PQ=2 ∣k∣P=(k,0),\;\; Q=(0,k),\;\; PQ=\sqrt2\,|k|

A Square Inscribed in a Circle

Inscribed-square vertices

(h±r2,  k±r2)\left(h\pm\tfrac{r}{\sqrt2},\; k\pm\tfrac{r}{\sqrt2}\right)

Tangent and Normal at a Point of Contact

Opposite end of the diameter

T′=2C−T(C=centre,  T=contact point)T'=2C-T\quad(C=\text{centre},\;T=\text{contact point})

Areas of the Minor and Major Segments

Minor segment area

Aminor=a22 (θ−sin⁡θ)A_{\text{minor}} = \tfrac{a^2}{2}\,(\theta - \sin\theta)
  • aaradius
  • θ\thetacentral angle (radians)

Common traps

Don't forget the supplementary (obtuse) case

The inscribed angle depends on which arc AA is on: 12∠BOC\tfrac12\angle BOC on the major arc, its supplement on the minor arc. NDA answer keys frequently list BOTH π/4\pi/4 and 3π/43\pi/4. Quoting only the acute value loses the obtuse option.

A is not pinned to one coordinate

If a question asks for "the coordinates of AA" with only B,CB,C given, there are infinitely many valid points on the arc — the answer is a locus, not a single point. Watch for the choice that says "cannot be uniquely determined".

The contact point shares ONE coordinate with the centre

On the x-axis the contact point is (k,0)(k,0) — same x as the centre, y zero — because the radius to the contact point is vertical. Reading it as (0,k)(0,k) or (k,k)(k,k) is the usual mix-up.

Inscribed vs circumscribed — offset is r/√2, not r

An INSCRIBED square (corners on the circle) has vertices offset r/2r/\sqrt2 from the centre. A CIRCUMSCRIBED square (sides tangent to the circle) has vertices offset rr. Mixing the two gives (h±r,k±r)(h\pm r,k\pm r) — the wrong, larger square.

The normal goes through the centre — that's the whole trick

The normal at a circle's point is the radius line, so it always passes through the centre. The far intersection with the circle is the diametrically opposite point, 2C−T2C-T. Trying to solve the normal–circle intersection from scratch wastes time and invites sign errors.

Segment = sector − triangle (not sector alone)

The minor SEGMENT is the sector with the triangle cut off: a22(θ−sin⁡θ)\tfrac{a^2}{2}(\theta-\sin\theta). Forgetting the −sin⁡θ-\sin\theta term reports the SECTOR area instead — a different region. The major segment is then the whole disc minus the minor segment, not 'the big sector'.

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