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Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

If the two circles $${(x - 1)^2}\, + \,{(y - 3)^2} = \,{r^2}$$ and $$\,{x^2}\, + \,{y^2} - \,8x\, + \,2y\, + \,\,8\,\, = 0$$ intersect in two distinct point, then

A

$$r > 2$$

B

$$2 < r < 8$$

C

$$r < 2$$

D

$$r = 2.$$

$$\left| {{r_1} - {r_2}} \right| < {C_1}{C_2}$$ for intersection

$$ \Rightarrow r - 3 < 5 \Rightarrow r < 8\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

and $${r_1} + {r_2} > {C_1}{C_2},\,$$

$$r + 3 > 5 \Rightarrow r > 2\,\,\,...\left( 2 \right)$$

From $$\left( 1 \right)$$ and $$\left( 2 \right),$$ $$2 < r < 8.$$

$$ \Rightarrow r - 3 < 5 \Rightarrow r < 8\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

and $${r_1} + {r_2} > {C_1}{C_2},\,$$

$$r + 3 > 5 \Rightarrow r > 2\,\,\,...\left( 2 \right)$$

From $$\left( 1 \right)$$ and $$\left( 2 \right),$$ $$2 < r < 8.$$

2

MCQ (Single Correct Answer)

The centres of a set of circles, each of radius 3, lie on the circle $${x^2}\, + \,{y^2} = 25$$. The locus of any point in the set is

A

$$4\, \le \,\,{x^2}\, + \,{y^2}\, \le \,\,64$$

B

$${x^2}\, + \,{y^2}\, \le \,\,25$$

C

$${x^2}\, + \,{y^2}\, \ge \,\,25$$

D

$$3\, \le \,\,{x^2}\, + \,{y^2}\, \le \,\,9$$

For any point $$P(x,y)$$ in the given circle,

we should have

$$OA \le OP \le OB$$

$$ \Rightarrow \left( {5 - 3} \right) \le \sqrt {{x^2} + {y^2}} \le 5 + 3$$

$$ \Rightarrow 4 \le {x^2} + {y^2} \le 64$$

we should have

$$OA \le OP \le OB$$

$$ \Rightarrow \left( {5 - 3} \right) \le \sqrt {{x^2} + {y^2}} \le 5 + 3$$

$$ \Rightarrow 4 \le {x^2} + {y^2} \le 64$$

3

MCQ (Single Correct Answer)

The equation of a circle with origin as a center and passing thorough equilateral triangle whose median is of length $$3$$ $$a$$ is

A

$${x^2}\, + \,{y^2} = 9{a^2}$$

B

$${x^2}\, + \,{y^2} = 16{a^2}$$

C

$${x^2}\, + \,{y^2} = 4{a^2}$$

D

$${x^2}\, + \,{y^2} = {a^2}$$

Let $$ABC$$ be an equilateral triangle, whose median is $$AD.$$

Given $$AD=3a.$$

In $$\Delta ABD,\,\,A{B^2} = A{D^2} + B{D^2};$$

$$ \Rightarrow {x^2} = 9{a^2} + \left( {{x^2}/4} \right)\,\,$$

where $$AB = BC = AC = x.$$

$${3 \over 4}{x^2} = 9{a^2} \Rightarrow {x^2} = 12{a^2}.$$

In $$\,\,\,\Delta OBD,O{B^2} = O{D^2} + B{D^2}$$

$$ \Rightarrow {r^2} = {\left( {3a - r} \right)^2} + {{{x^2}} \over 4}$$

$$ \Rightarrow {r^2} = 9{a^2} - 6ar + {r^2} + 3{a^2};$$

$$ \Rightarrow 6ar = 12{a^2}$$

$$ \Rightarrow r = 2a$$

So equation of circle is $${x^2} + {y^2} = 4{a^2}$$

Given $$AD=3a.$$

In $$\Delta ABD,\,\,A{B^2} = A{D^2} + B{D^2};$$

$$ \Rightarrow {x^2} = 9{a^2} + \left( {{x^2}/4} \right)\,\,$$

where $$AB = BC = AC = x.$$

$${3 \over 4}{x^2} = 9{a^2} \Rightarrow {x^2} = 12{a^2}.$$

In $$\,\,\,\Delta OBD,O{B^2} = O{D^2} + B{D^2}$$

$$ \Rightarrow {r^2} = {\left( {3a - r} \right)^2} + {{{x^2}} \over 4}$$

$$ \Rightarrow {r^2} = 9{a^2} - 6ar + {r^2} + 3{a^2};$$

$$ \Rightarrow 6ar = 12{a^2}$$

$$ \Rightarrow r = 2a$$

So equation of circle is $${x^2} + {y^2} = 4{a^2}$$

4

MCQ (Single Correct Answer)

The centre of the circle passing through (0, 0) and (1, 0) and touching the circle $${x^2}\, + \,{y^2} = 9$$ is

A

$$\left( {{1 \over 2},\,{1 \over 2}} \right)$$

B

$$\left( {{1 \over 2},\, - \,\sqrt 2 } \right)$$

C

$$\left( {{3 \over 2},\,{1 \over 2}} \right)$$

D

$$\left( {{1 \over 2},\,{3 \over 2}} \right)$$

Let the required circle be

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

Since it passes through $$\left( {0,0} \right)$$ and $$\left( {1,0} \right)$$

$$ \Rightarrow c = 0$$ and $$g = - {1 \over 2}$$

Points $$\left( {0,0} \right)$$ and $$\left( {1,0} \right)$$ lie inside the circle $${x^2} + {y^2} = 9,$$

so two circles touch internally

$$ \Rightarrow c{}_1{c_2} = {r_1} - {r_2}$$

$$\therefore$$ $$\sqrt {{g^2} + {f^2}} = 3 - \sqrt {{g^2} + {f^2}} $$

$$ \Rightarrow \sqrt {{g^2} + {f^2}} = {3 \over 2}$$

$$ \Rightarrow {f^2} = {9 \over 4} - {1 \over 4} = 2$$

$$\therefore$$ $$f = \pm \sqrt 2 .$$

Hence, the centers of required circle are

$$\left( {{1 \over 2}.\sqrt 2 } \right)$$ or $$\left( {{1 \over 2}, - \sqrt 2 } \right)$$

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

Since it passes through $$\left( {0,0} \right)$$ and $$\left( {1,0} \right)$$

$$ \Rightarrow c = 0$$ and $$g = - {1 \over 2}$$

Points $$\left( {0,0} \right)$$ and $$\left( {1,0} \right)$$ lie inside the circle $${x^2} + {y^2} = 9,$$

so two circles touch internally

$$ \Rightarrow c{}_1{c_2} = {r_1} - {r_2}$$

$$\therefore$$ $$\sqrt {{g^2} + {f^2}} = 3 - \sqrt {{g^2} + {f^2}} $$

$$ \Rightarrow \sqrt {{g^2} + {f^2}} = {3 \over 2}$$

$$ \Rightarrow {f^2} = {9 \over 4} - {1 \over 4} = 2$$

$$\therefore$$ $$f = \pm \sqrt 2 .$$

Hence, the centers of required circle are

$$\left( {{1 \over 2}.\sqrt 2 } \right)$$ or $$\left( {{1 \over 2}, - \sqrt 2 } \right)$$

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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

Trigonometric Functions & Equations

Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

Functions

Limits, Continuity and Differentiability

Differentiation

Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations