JEE Exam » JEE Important Formulas » JEE Maths Formulas – Part 1

JEE Maths Formulas – Part 1

In this article, we will go through maths quick formula revision for JEE 2023. Find the important formulas for circle, quadratic equation and binomial theorem.

Circle Formula

The formula for circle are as stated below



Area of a Circle

  • In terms of radius: πr^2

  • In terms of diameter:  π/4×d^2

Surface Area of a Circle


General Equation of a Circle

The general equation of a circle with coordinates of a centre(h,k), and radius r is given as:√((〖x-h)〗^2+(〖y-k)〗^2 )=r

Standard Equation of a Circle

The Standard equation of a circle with centre (a,b), and radius r is given as: (〖x-a)〗^2+(〖y-b)〗^2=r^2

Diameter of a Circle

2 radius

Circumference of a Circle


Intercepts made by Circle


  1. On x-axis: 2√(g^2-c)

  2. On y-axis: 2√(f^2-c)

Parametric Equations of a Circle

x=h+rcos θ ;y=k+rsin θ


  • Slope form : y=mx±a√(1+m^2 )

  • Point form: xx_1+yy_1=a^2

  • Parametric form: xcos α +ysin α =a

Pair of Tangents from a Point:


Length of a Tangent

√(S_1 )

Director Circle


Chord of Contact


  1. Length of chord of contact= 2LR/√(R^2+L^2 )

  2. Area of the triangle formed by the pair of the tangents and its chord of contact = (RL^3)/(R^2+L^2 )

  3. Tangent of the angle between the pair of tangents from (x_1,y_1 )=(2RL/(L^2-R^2 ))

  4. Equation of the circle circumscribing the triangle PT_1,T_2  is:                                     (x-x_1 )(x+g)+(y-y_1 )(y+f)=0

Condition of orthogonality of Two Circles

2g_1 g_2+2f_1 f_2=c_1+c_2

Radical Axis

S_1-S_2=0 i.e. 2(g_1-g_2 )x+2(f_1-f_2 )y+(c_1-c_2 )=0

Family of Circles


Quadratic Equation Formula

The formula for quadratic equation are as stated below



General form of Quadratic Equation

where a,b,c are constants and a≠0.

Roots of equations

α=(-b+√(b^2-4ac))/2a  β=- (-b-√(b^2-4ac))/2a

Sum and Product of Roots

If and are the roots of the quadratic equation ax^2+bx+c =0, then

Sum of roots, α+β=-b/a

Product of roots, αβ=c/a

Discriminant of Quadratic equation

The Discriminant of the quadratic equation is ax^2+bx+c=0 =0 given by D=b^2-4ac

Nature of Roots

  • If D=0, the roots are real and equalαβ=-b/2a

  • If D≠0, The roots are real and unequal

  • If D<0, the roots are imaginary and unequal

  • If D>0 and D is a perfect square, the roots are rational and unequal

  • If D>0 and D is not a perfect square, the roots are irrational and unequal

Formation of Quadratic Equation with given roots

If α and are the roots of the quadratic equation, thenx-αx-β=0; x^2-(α+β)x+αβ=0

  • x^2-(Sum of roots)x+product of roots=0

Common Roots

  • If two quadratic equations  a_2 x^2+b_2 x+c_2=0 &   have both roots common, then a_1/a_2 =b_1/b_2 =c_1/c_2

  • If only one root α is common, then α=(c_1 a_2-c_2 a_1)/(a_1 b_2-a_2 b_1 )=(b_1 c_2-b_2 c_1)/(c_1 a_2-c_2 a_1 )

Range of Quadratic Expression fx=ax2+bx+c in restricted domain x∈[x1,x2]

  • If b/2a not belong to [x1,x2] then, f(x)∈fx1,f(x2) ,  max⁡{fx1,f(x2)}
  • Ifb/2a∈[x1,x2] then, f(x)∈fx1,fx2,-D/4a ,  max⁡{fx1,fx2,-D/4a}

Roots under special cases

Consider the quadratic equation ax^2+bx+c=0

  • If c=0, then one root is zero. Other root is-b/a

  • If b=0The roots are equal but in opposite signs

  • If b=c=0, then both roots are zero

  • If a=c, then the roots are reciprocal to each other

  • If a+b+c=0, then one root is 1 and the second root is ca

  • If a=b=c=0, then the equation will become an identity and will satisfy every value of x

Graph of Quadratic equation

The graph of a quadratic equation ax^2+bx+c=0 is a parabola.

  • If a>0, then the graph of a quadratic equation will be concave upwards

  • If a<0, then the graph of a quadratic equation will be concave downwards

Maximum and Minimum value

Consider the quadratic expression ax^2+bx+c=0

  • If a<0, then the expression has the greatest value at x=-b/2a

  • The maximum value is -D/4a

  • If a>0, then the expression has the least value at x=-b/2a

  • The minimum value is -D4a

Quadratic Expression in Two Variables

The general form of a quadratic equation in two variables x and y is ax^2+2hxy+by^2+2gx+2fy+c

To solve the expression into two linear rational factors, the condition is ∆=0

       [ a  h  g ]

∆= [ h  b  f ] =0

        [ g  f  c ]

 abc+2fgh-af^2-bg^2-ch^2=0And 〖 h〗^2-ab>0. This is called the Discriminant of the given expression.

Binomial Theorem Formula

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Binomial Theorem for positive Integral Index

(x+a)^n=nC0x^n a^0+ nC1x^(n-1) a+ nC2x^(n-2) a^2+⋯+ nCrx^(n-r) a^r+⋯+ nCn.xa^n

General terms = T_(r+1)=nCrx^(n-r) a^r

Deductions of Binomial Theorem

(1+x)^n=nC0+ nC1x+ nC2x^2+nC3x^3+⋯+nCrx^r+⋯+nCnx^n

General Term=x^r=(n(n-1)(n-2)……(n-r+1))/r!.x^r

  • (1-x)^n=nC0-nC1x+nC2x^2-nC3x^3+⋯+(〖-1)〗^rnCrx^r+⋯+(-1)^nnCnx^n

General Term=x^r=(n(n-1)(n-2)……(n-r+1))/r!.x^r

Middle Term in the expansion ofx+an

  • If n is even then middle (n/2+1)^th

  • If n is odd then middle terms are ((n+1)/2)and (n+3)/2 term.

  • Binomial coefficients of middle term is the greatest Binomial coefficients

To determine a particular term in the expansion

In the expansion of(x^α±1/x^β )^n
, if xm occurs in 〖 T〗_(r+1), then r is given by  nα-r(α+β)=m=> r=(nα-m)/(α+β)and the term which is independent of x then

 nα-r(α+β)=0 => r=nα/(α+β)

To find a term from the end in the expansion of x+an

T_r (E)=T_(n-r+2) (B)

Binomial Coefficients and their properties

In the expansion of(1+x)^n=C_0+C_1 x+C_2 x^2+⋯+C_r x^r+⋯+C_n x^n

Where C_0=1,C_1=n,C_2=(n(n-1))/2!








Greatest term in the expansion of x+an:

  • The term in the expansion of (x+a)^nof greatest coefficients ={T n+2/2 When n is even T_(((n+1))/2),T_(((n+3))/2)  when n is odd
  • The greatest term=T_p ,T_(p+1) when
  • (n+1)a/(x+a)=p∈Z T_(q+1, )

 nnot belong to q<(n+1)a/(x+a) <q+1

Multinomial Expansion

If n∈N then the general terms of multinomial expansion

(x_1+x_2+x_3+⋯+x_k )^n   is ∑_(r_1+r_2+⋯+r_k=n) n!/(r_1 !r_2 !…r_k !) x_1^r1.x_2^r2…x_k^(r_k )


Binomial Theorem for Negative Integer Or Fractional Indices

(1+x)^n=1+nx+(n(n-1))/2! x^2+(n(n-1)(n-2))/3! x^3+⋯ +(n(n-1)(n-2)…….(n-r+1))/r! x^r+⋯,|x|<1T_(r+1)=(n(n-1)(n-2)…….(n-r+1))/r! x^r