JEE Maths Formulas – Part 1

Circle Formula

The formula for circle are as stated below

Description 

Formula 

Area of a Circle

• In terms of radius: $πr^2$

• In terms of diameter:  $π/4×d^2$

Surface Area of a Circle

$πr^2$

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

2πr

Intercepts made by Circle

$x^2+y^2+2gx+2fy+c=0$

1. $On x-axis: 2√(g^2-c)$

2. $On y-axis: 2√(f^2-c)$

Parametric Equations of a Circle

$x=h+rcos θ ;y=k+rsin θ$

Tangent

• 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:

$SS_1=T^2$

Length of a Tangent

$√(S_1 )$

Director Circle

$x^2+y^2=2a^2$

Chord of Contact

T=0

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

$S_1+KS_2=0,S+KL=0$

Quadratic Equation Formula

The formula for quadratic equation are as stated below

Description

Formula

General form of Quadratic Equation

 $ax^2+bx+c=0$
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)}
• If–b/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 ]

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

Binomial Theorem Formula

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Description

Formula

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

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

$C_0+C_1+C_2+……+C_n=2^n$

$C_0-C_1+C_2-C_3+……=0$

$C_0+C_2+……=C_1+C_3+……=2^(n-1)$

$C_0^2+C_1^2+C_2^2+……+C_n^2=2n!/n!n!$

$C_0+C_1/2+C_2/3+……+C_n/(n+1)=(2^(n+1)-1)/(n+1)$

$C_0-C_1/2+C_2/3-C_3/4+……+((-1)^n.C_n)/(n+1)=1/(n+1)$

Greatest term in the expansion of x+an:

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