Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

Let $$C$$ be the circle with centre at $$(1, 1)$$ and radius $$=$$ $$1$$. If $$T$$ is the circle centred at $$(0, y)$$, passing through origin and touching the circle $$C$$ externally, then the radius of $$T$$ is equal to

A

$${1 \over 2}$$

B

$${1 \over 4}$$

C

$${{\sqrt 3 } \over {\sqrt 2 }}$$

D

$${{\sqrt 3 } \over 2}$$

Equation of circle $$C \equiv {\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} = 1$$

Radius of $$T = \left| y \right|$$

$$T$$ touches $$C$$ externally

therefore,

Distance between the centers $$=$$ sum of their radii

$$ \Rightarrow \sqrt {{{\left( {0 - 1} \right)}^2} + {{\left( {y - 1} \right)}^2}} = 1 + \left| y \right|$$

$$ \Rightarrow {\left( {0 - 1} \right)^2} + {\left( {y - 1} \right)^2} = {\left( {1 + \left| y \right|} \right)^2}$$

$$ \Rightarrow 1 + {y^2} + 1 - 2y = 1 + {y^2} + 2\left| y \right|$$

$$2\left| y \right| = 1 - 2y$$

If $$y>0$$ then $$2y=1-2y$$ $$ \Rightarrow y = {1 \over 4}$$

$$y<0$$ then $$-2y=1-2y$$ $$ \Rightarrow 0 = 1$$ (not possible)

$$\therefore$$ $$y = {1 \over 4}$$

2

MCQ (Single Correct Answer)

The circle passing through $$(1, -2)$$ and touching the axis of $$x$$ at $$(3, 0)$$ also passes through the point

A

$$\left( { - 5,\,2} \right)$$

B

$$\left( { 2,\,-5} \right)$$

C

$$\left( { 5,\,-2} \right)$$

D

$$\left( { - 2,\,5} \right)$$

Since circle touches $$x$$-axis at $$(3,0)$$

$$\therefore$$ The equation of circle be

$${\left( {x - 3} \right)^2} + {\left( {y - 0} \right)^2} + \lambda y = 0$$

As it passes through $$(1, -2)$$

$$\therefore$$ Put $$x=1,$$ $$y=-2$$

$$ \Rightarrow {\left( {1 - 3} \right)^2} + {\left( { - 2} \right)^2} + \lambda \left( { - 2} \right) = 0$$

$$ \Rightarrow \lambda = 4$$

$$\therefore$$ equation of circle is $${\left( {x - 3} \right)^2} + {y^2} - 8 = 0$$

Now, from the options $$\left( {5, - 2} \right)$$ satisfies equation of circle.

$$\therefore$$ The equation of circle be

$${\left( {x - 3} \right)^2} + {\left( {y - 0} \right)^2} + \lambda y = 0$$

As it passes through $$(1, -2)$$

$$\therefore$$ Put $$x=1,$$ $$y=-2$$

$$ \Rightarrow {\left( {1 - 3} \right)^2} + {\left( { - 2} \right)^2} + \lambda \left( { - 2} \right) = 0$$

$$ \Rightarrow \lambda = 4$$

$$\therefore$$ equation of circle is $${\left( {x - 3} \right)^2} + {y^2} - 8 = 0$$

Now, from the options $$\left( {5, - 2} \right)$$ satisfies equation of circle.

3

MCQ (Single Correct Answer)

The length of the diameter of the circle which touches the $$x$$-axis at the point $$(1, 0)$$ and passes through the point $$(2, 3)$$ is:

A

$${{10} \over 3}$$

B

$${{3} \over 5}$$

C

$${{6} \over 5}$$

D

$${{5} \over 3}$$

Let center of the circle be $$\left( {1,h} \right)$$

$$\left[ {\,\,} \right.$$ as circle touches $$x$$-axis at $$\left. {\left( {1,0} \right)\,\,} \right]$$

Let the circle passes through the point $$B(2,3)$$

$$\therefore$$ $$CA=CB$$ (radius)

$$ \Rightarrow C{A^2} = C{B^2}$$

$$ \Rightarrow {\left( {1 - 1} \right)^2} + \left( {h - 0} \right){}^2 = {\left( {1 - 2} \right)^2} + {\left( {h - 3} \right)^2}$$

$$ \Rightarrow {h^2} = 1 + {h^2} + 9 - 6h$$

$$ \Rightarrow h = {{10} \over 6} = {5 \over 3}$$

Thus, diameter is $$2h = {{10} \over 3}.$$

$$\left[ {\,\,} \right.$$ as circle touches $$x$$-axis at $$\left. {\left( {1,0} \right)\,\,} \right]$$

Let the circle passes through the point $$B(2,3)$$

$$\therefore$$ $$CA=CB$$ (radius)

$$ \Rightarrow C{A^2} = C{B^2}$$

$$ \Rightarrow {\left( {1 - 1} \right)^2} + \left( {h - 0} \right){}^2 = {\left( {1 - 2} \right)^2} + {\left( {h - 3} \right)^2}$$

$$ \Rightarrow {h^2} = 1 + {h^2} + 9 - 6h$$

$$ \Rightarrow h = {{10} \over 6} = {5 \over 3}$$

Thus, diameter is $$2h = {{10} \over 3}.$$

4

MCQ (Single Correct Answer)

The two circles x^{2} + y^{2} = ax, and x^{2} + y^{2} = c^{2} (c > 0) touch each other if

A

| a | = c

B

a = 2c

C

| a | = 2c

D

2 | a | = c

As center of one circle is $$\left( {0,0} \right)$$ and other circle passes through $$(0,0),$$ therefore

Also $${C_1}\left( {{a \over 2},0} \right){C_2}\left( {0,0} \right)$$

$${r_1} = {a \over 2}{r_2} = C$$

$${C_1}{C_2} = {r_1} - {r_2} = {a \over 2}$$

$$ \Rightarrow C - {a \over 2} = {a \over 2}$$

$$ \Rightarrow C = a$$

If the two circles touch each other, then they must touch each other internally.

Also $${C_1}\left( {{a \over 2},0} \right){C_2}\left( {0,0} \right)$$

$${r_1} = {a \over 2}{r_2} = C$$

$${C_1}{C_2} = {r_1} - {r_2} = {a \over 2}$$

$$ \Rightarrow C - {a \over 2} = {a \over 2}$$

$$ \Rightarrow C = a$$

If the two circles touch each other, then they must touch each other internally.

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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