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MHT-CET Maths · Formula sheet

Circle formulas

15 formulas and 15 common traps for MHT-CET Maths Circle, grouped by subtopic.

Full notes with worked examples

Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms

Learn this subtopic in the notes

Centre-Radius and General Forms: Read (−g, −f) and √(g² + f² − c)

Two forms

(x−h)2+(y−k)2=r2x2+y2+2gx+2fy+c=0: C(−g,−f), r=g2+f2−c(x - h)^2 + (y - k)^2 = r^2 \qquad x^2 + y^2 + 2gx + 2fy + c = 0:\ C(-g, -f),\ r = \sqrt{g^2 + f^2 - c}

Diameter Form: (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0

Diameter form

(x−x1)(x−x2)+(y−y1)(y−y2)=0  ⟺  x2+y2−(x1+x2)x−(y1+y2)y+x1x2+y1y2=0(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \iff x^2 + y^2 - (x_1 + x_2)x - (y_1 + y_2)y + x_1x_2 + y_1y_2 = 0

Parametric Form: x = h + r cos θ, y = k + r sin θ

Parametric circle

x=h+rcos⁡θ,y=k+rsin⁡θ(0≤θ<2π)x = h + r\cos\theta,\quad y = k + r\sin\theta \qquad (0 \le \theta < 2\pi)

Points of a Family on a Circle: Substitute, Get a Polynomial, Use Vieta

Vieta on the substituted polynomial

m4+2gm3+cm2+2fm+1=0 ⇒ m1m2m3m4=1m^4 + 2gm^3 + cm^2 + 2fm + 1 = 0 \ \Rightarrow\ m_1 m_2 m_3 m_4 = 1

Common traps

Reading the centre as (g, f)

x2+y2−4x+6y−3=0x^2 + y^2 - 4x + 6y - 3 = 0 has centre (2,−3)(2, -3): the signs FLIP. Both sign-error centres are always on the option list.

Sign of the constant with negative products

x1x2+y1y2=−b2−q2x_1x_2 + y_1y_2 = -b^2 - q^2. The option with −b2+q2-b^2 + q^2 is the planted slip; the paper prints −(b2+q2)-(b^2 + q^2).

Halving the radius with the centre

For x2+y2−ax−by=0x^2 + y^2 - ax - by = 0 the centre halves a,ba, b and the radius is a2+b22\dfrac{\sqrt{a^2 + b^2}}{2}, not a2+b24\dfrac{\sqrt{a^2 + b^2}}{4}. Options (B) and (D) carry the quartered radius.

Product of roots with the wrong sign

For an even-degree monic polynomial the product of the roots is ++(constant term); −1-1 is option (A) and comes from applying the cubic's sign rule.

Concentric Circles and Circles Touching a Line or an Axis

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Concentric Circles: Same Centre, New Radius From Area or From a Point

Concentric family

x2+y2+2gx+2fy+λ=0double area: R2=2r2x^2 + y^2 + 2gx + 2fy + \lambda = 0 \qquad \text{double area: } R^2 = 2r^2

Touching a Line or an Axis: Radius = Distance From the Centre; Contact Point = Foot of the Perpendicular

Tangency of a line

∣ah+bk+c∣a2+b2=rtouches the X-axis  ⟺  r=∣k∣\frac{|ah + bk + c|}{\sqrt{a^2 + b^2}} = r \qquad \text{touches the } X\text{-axis} \iff r = |k|

Common traps

Doubling the radius for double the area

Double AREA means R=2 rR = \sqrt2\, r, R2=2r2R^2 = 2r^2. Doubling rr quadruples the area and gives x2+y2−6x−4y=87x^2 + y^2 - 6x - 4y = 87, which is not on the list — but 5050 (the value of R2R^2, not the constant) is.

Using the centre's x-coordinate for tangency to the X-axis

Touching the XX-axis fixes r=∣k∣r = |k|, the yy-coordinate. With centre (3,2)(3, 2), r=2r = 2, constant 99; using r=3r = 3 gives constant 44, option (D).

Tangents — At a Point, With a Given Slope, From an External Point and Their Loci

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Tangent at a Point on the Circle: T = 0

Tangent at a point

xx1+yy1+g(x+x1)+f(y+y1)+c=0x2+y2=a2: xcos⁡θ+ysin⁡θ=axx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0 \qquad x^2 + y^2 = a^2:\ x\cos\theta + y\sin\theta = a

Tangent of a Given Slope, and Whether a Line Touches: Distance From the Centre = Radius

Tangency condition

y=mx±a1+m2general: ∣ah+bk+c∣a2+b2=ry = mx \pm a\sqrt{1 + m^2} \qquad \text{general: } \frac{|ah + bk + c|}{\sqrt{a^2 + b^2}} = r

Tangents From an External Point: Length √S₁, the Kite, the Angle Between Them

External point

L=S1=x12+y12+2gx1+2fy1+c,[PAOB]=rL,sin⁡θ2=rCPL = \sqrt{S_1} = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c},\qquad [PAOB] = rL,\qquad \sin\frac{\theta}{2} = \frac{r}{CP}

Loci From Tangent Lengths, and PQ · RS = (2r)²

Tangent-length locus

S1S2=mn  ⟺  n2S1=m2S2PQ⋅RS=(2r)2\frac{\sqrt{S_1}}{\sqrt{S_2}} = \frac{m}{n} \iff n^2 S_1 = m^2 S_2 \qquad PQ \cdot RS = (2r)^2

Common traps

Forgetting to halve the linear coefficients in T

−6x-6x becomes −3(x+x1)-3(x + x_1), not −6(x+x1)-6(x + x_1). Doubling gives 8x−2y−52=08x - 2y - 52 = 0-style options that are exactly the planted distractors.

Using the slope of the given line instead of the perpendicular one

'Perpendicular to 5x+y=25x + y = 2' means slope 15\dfrac15, giving x−5y±626=0x - 5y \pm 6\sqrt{26} = 0. Options (C) and (D) use slope −5-5 or 55.

Halving the kite

PAOBPAOB is TWO right triangles, so its area is rLrL, not 12rL\tfrac12 rL. 232\sqrt3 is option (A) on the (−4,0)(-4, 0) stem; the answer is 434\sqrt3.

Sign of the x term after cross-multiplying

9(4x)−4(−6x)=36x+24x=+60x9(4x) - 4(-6x) = 36x + 24x = +60x. The bank once carried −60x-60x as its key; the sign is positive.

Distance From a Point to a Circle — Greatest, Least, a Line Cutting the Circle and the Segment Area

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Greatest and Least Distance: d ± r From a Point, and From the Circle to a Line

Extreme distances

min⁡=∣d−r∣,max⁡=d+r(d=distance from the point, or the line, to the centre)\min = |d - r|,\quad \max = d + r \qquad (d = \text{distance from the point, or the line, to the centre})

When a Line Cuts the Circle: Distance From the Centre < r

Line and circle

p=∣ah+bk+c∣a2+b2:p<r cuts, p=r touches, p>r misses;chord=2r2−p2p = \frac{|ah + bk + c|}{\sqrt{a^2 + b^2}}:\quad p < r \text{ cuts},\ p = r \text{ touches},\ p > r \text{ misses};\qquad \text{chord} = 2\sqrt{r^2 - p^2}

Area Cut Off by a Chord, and the Circumcircle of an Equilateral Triangle

Segment and circumradius

segment=r22(θ−sin⁡θ)equilateral: median=3R2\text{segment} = \frac{r^2}{2}(\theta - \sin\theta) \qquad \text{equilateral: median} = \frac{3R}{2}

Common traps

Adding r twice for the far end of the diameter

AM′=AM+2rAM' = AM + 2r because MM′MM' is a whole diameter — 5+10=155 + 10 = 15. AM+r=10AM + r = 10 is option (A).

Counting the tangent cases

'Two distinct points' is strict: m=−8m = -8 and m=2m = 2 are tangents. Including them gives 1111, option (D); excluding both gives 99.

Taking R as the median

The circumradius is two-thirds of the median, so the median is 32R=35\dfrac{3}{2}R = 3\sqrt5; 252\sqrt5 itself is option (A).

Two Circles — Touching, Common Tangents and Relative Position

Learn this subtopic in the notes

Relative Position From d, r₁ + r₂ and |r₁ − r₂|: How Many Common Tangents

Five positions

d>r1+r2: 4d=r1+r2: 3∣r1−r2∣<d<r1+r2: 2d=∣r1−r2∣: 1d<∣r1−r2∣: 0d > r_1 + r_2:\ 4 \quad d = r_1 + r_2:\ 3 \quad |r_1 - r_2| < d < r_1 + r_2:\ 2 \quad d = |r_1 - r_2|:\ 1 \quad d < |r_1 - r_2|:\ 0

Touching Circles: d = r₁ + r₂ (External) or d = |r₁ − r₂| (Internal), and the Centre From the Contact Point

Touching

external: d=r1+r2internal: d=∣r1−r2∣centres and contact point are collinear\text{external: } d = r_1 + r_2 \qquad \text{internal: } d = |r_1 - r_2| \qquad \text{centres and contact point are collinear}

Common traps

Comparing d with r₁ + r₂ only

d<r1+r2d < r_1 + r_2 covers three positions (cutting, internal touch, nested). The second comparison, against ∣r1−r2∣|r_1 - r_2|, decides between 22, 11 and 00.

Squaring once and stopping

a2+b2=a2−c+b2−c\sqrt{a^2 + b^2} = \sqrt{a^2 - c} + \sqrt{b^2 - c} needs TWO squarings; after the first, 2c=2(a2−c)(b2−c)2c = 2\sqrt{(a^2 - c)(b^2 - c)} still has a root. The half-done version gives 1a2+1b2=1c2\dfrac{1}{a^2} + \dfrac{1}{b^2} = \dfrac{1}{c^2}, option (C).

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