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1

MCQ (Single Correct Answer)

Let $$C$$ be the circle with centre $$(0, 0)$$ and radius $$3$$ units. The equation of the locus of the mid points of the chords of the circle $$C$$ that subtend an angle of $${{2\pi } \over 3}$$ at its center is

A

$${x^2} + {y^2} = {3 \over 2}$$

B

$${x^2} + {y^2} = 1$$

C

$${x^2} + {y^2} = {{27} \over 4}$$

D

$${x^2} + {y^2} = {{9} \over 4}$$

Let $$M\left( {h,k} \right)$$ be the mid point of chord $$AB$$ where

$$\angle AOB = {{2\pi } \over 3}$$

$$\therefore$$ $$\angle AOM = {\pi \over 3}.$$ Also $$OM=$$ $$3\cos {\pi \over 3} = {3 \over 2}$$

$$ \Rightarrow \sqrt {{h^2} + k{}^2} = {3 \over 2}$$

$$ \Rightarrow {h^2} + {k^2} = {9 \over 4}$$

$$\therefore$$ Locus of $$\left( {h,k} \right)$$ is

$${x^2} + {y^2} = {9 \over 4}$$

$$\angle AOB = {{2\pi } \over 3}$$

$$\therefore$$ $$\angle AOM = {\pi \over 3}.$$ Also $$OM=$$ $$3\cos {\pi \over 3} = {3 \over 2}$$

$$ \Rightarrow \sqrt {{h^2} + k{}^2} = {3 \over 2}$$

$$ \Rightarrow {h^2} + {k^2} = {9 \over 4}$$

$$\therefore$$ Locus of $$\left( {h,k} \right)$$ is

$${x^2} + {y^2} = {9 \over 4}$$

2

MCQ (Single Correct Answer)

If the lines $$3x - 4y - 7 = 0$$ and $$2x - 3y - 5 = 0$$ are two diameters of a circle of area $$49\pi $$ square units, the equation of the circle is

A

$$\,{x^2} + {y^2} + 2x\, - 2y - 47 = 0\,$$

B

$$\,{x^2} + {y^2} + 2x\, - 2y - 62 = 0\,$$

C

$${x^2} + {y^2} - 2x\, + 2y - 62 = 0$$

D

$${x^2} + {y^2} - 2x\, + 2y - 47 = 0$$

Point of intersection of $$3x - 4y - 7 = 0$$ and

$$2x - 3y - 5 = 0$$ is $$\left( {1, - 1} \right)$$ which is the center of the

circle and radius $$=7$$

$$\therefore$$ Equation is $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 49$$

$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 47 = 0$$

$$2x - 3y - 5 = 0$$ is $$\left( {1, - 1} \right)$$ which is the center of the

circle and radius $$=7$$

$$\therefore$$ Equation is $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 49$$

$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 47 = 0$$

3

MCQ (Single Correct Answer)

If the pair of lines $$a{x^2} + 2\left( {a + b} \right)xy + b{y^2} = 0$$ lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then

A

$$3{a^2} - 10ab + 3{b^2} = 0$$

B

$$3{a^2} - 2ab + 3{b^2} = 0$$

C

$$3{a^2} + 10ab + 3{b^2} = 0$$

D

$$3{a^2} + 2ab + 3{b^2} = 0$$

As per question area of one sector $$=3$$ area of another sector

$$ \Rightarrow $$ at center by one sector $$ = 3 \times $$ angle at center by another sector

Let one angle be $$\theta $$ then other $$=30$$

Clearly $$\theta + 3\theta = 180 \Rightarrow \theta = {45^ \circ }$$

$$\therefore$$ Angle between the diameters represented by combined equation

$$a{x^2} + 2\left( {a + b\,\,\,xy} \right) + b{y^2} = 0$$ is $${45^ \circ }$$

$$\therefore$$ Using $$tan$$ $$\theta $$ $$ = {{2\sqrt {{h^2} - ab} } \over {a + b}}$$

we get $$\tan \,{45^ \circ } = {{2\sqrt {{{\left( {a + b} \right)}^2} - ab} } \over {a + b}}$$

$$ \Rightarrow 1 = {{2\sqrt {{a^2} + {b^2} + ab} } \over {a + b}}$$

$$ \Rightarrow {\left( {a + b} \right)^2} = 4\left( {{a^2} + {b^2} + ab} \right)$$

$$ \Rightarrow {a^2} + {b^2} + 2ab = 4{a^2} + 4{b^2} + 4ab$$

$$ \Rightarrow 3{a^2} + 3{b^2} + 2ab = 0$$

4

MCQ (Single Correct Answer)

If the circles $${x^2}\, + \,{y^2} + \,2ax\, + \,cy\, + a\,\, = 0$$ and $${x^2}\, + \,{y^2} - \,3ax\, + \,dy\, - 1\,\, = 0$$ intersect in two ditinct points P and Q then the line 5x + by - a = 0 passes through P and Q for

A

exactly one value of a

B

no value of a

C

infinitely many values of a

D

exactly two values of a

$${s_1} = {x^2} + {y^2} + 2ax + cy + a = 0$$

$${s_2} = {x^2} + {y^2} - 3ax + dy - 1 = 0$$

Equation of common chord of circles $${s_1}$$ and $${s_2}$$ is

given by $${s_1} - {s_2} = 0$$

$$ \Rightarrow 5ax + \left( {c - d} \right)y + a + 1 = 0$$

Given that $$5x + by - a = 0$$ passes through $$P$$ and $$Q$$

$$\therefore$$ The two equations should represent the same line

$$ \Rightarrow {a \over 1} = {{c - d} \over b} = {{a + 1} \over { - a}}$$

$$ \Rightarrow a + 1 = - {a^2}$$

$${a^2} + a + 1 = 0$$

No real value of $$a.$$

$${s_2} = {x^2} + {y^2} - 3ax + dy - 1 = 0$$

Equation of common chord of circles $${s_1}$$ and $${s_2}$$ is

given by $${s_1} - {s_2} = 0$$

$$ \Rightarrow 5ax + \left( {c - d} \right)y + a + 1 = 0$$

Given that $$5x + by - a = 0$$ passes through $$P$$ and $$Q$$

$$\therefore$$ The two equations should represent the same line

$$ \Rightarrow {a \over 1} = {{c - d} \over b} = {{a + 1} \over { - a}}$$

$$ \Rightarrow a + 1 = - {a^2}$$

$${a^2} + a + 1 = 0$$

No real value of $$a.$$

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Complex Numbers

Quadratic Equation and Inequalities

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Sequences and Series

Matrices and Determinants

Vector Algebra and 3D Geometry

Probability

Statistics

Mathematical Reasoning

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Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

Functions

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Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations