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1

MCQ (Single Correct Answer)

The centres of those circles which touch the circle, $${x^2} + {y^2} - 8x - 8y - 4 = 0$$, externally and also touch the $$x$$-axis, lie on:

A

a hyperbola

B

a parabola

C

a circle

D

an ellipse which is not a circle

For the given circle,

center : $$(4,4)$$

radius $$=6$$

$$6 + k = \sqrt {{{\left( {h - 4} \right)}^2} + {{\left( {k - 4} \right)}^2}} $$

$${\left( {h - 4} \right)^2} = 20k + 20$$

$$\therefore$$ locus of $$(h, k)$$ is

$${\left( {h - 4} \right)^2} = 20\left( {y + 1} \right),$$

which is parabola.

2

MCQ (Single Correct Answer)

The number of common tangents to the circles $${x^2} + {y^2} - 4x - 6x - 12 = 0$$ and $${x^2} + {y^2} + 6x + 18y + 26 = 0,$$ is :

A

$$3$$

B

$$4$$

C

$$1$$

D

$$2$$

$${x^2} + {y^2} - 4x - 6y - 12 = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

Center, $${c_1} = \left( {2,\,3} \right)$$ and Radius, $${r_1} = 5$$ units

$${x^2} + {y^2} + 6x + 18y + 26 = 0\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Center, $${c_2} = \left( { - 3, - 9} \right)$$ and Radius, $${r_2} = 8$$ units

$${C_1}{C_2} = \sqrt {{{\left( {2 + 3} \right)}^2} + {{\left( {3 + 9} \right)}^2}} = 13\,\,$$ units

$${r_1} + {r_2} = 5 + 8 = 13$$

$$\therefore$$ $${C_1}{C_2} = {r_1} + {r_2}$$

Therefore there are three common tangents.

Center, $${c_1} = \left( {2,\,3} \right)$$ and Radius, $${r_1} = 5$$ units

$${x^2} + {y^2} + 6x + 18y + 26 = 0\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Center, $${c_2} = \left( { - 3, - 9} \right)$$ and Radius, $${r_2} = 8$$ units

$${C_1}{C_2} = \sqrt {{{\left( {2 + 3} \right)}^2} + {{\left( {3 + 9} \right)}^2}} = 13\,\,$$ units

$${r_1} + {r_2} = 5 + 8 = 13$$

$$\therefore$$ $${C_1}{C_2} = {r_1} + {r_2}$$

Therefore there are three common tangents.

3

MCQ (Single Correct Answer)

Locus of the image of the point $$(2, 3)$$ in the line $$\left( {2x - 3y + 4} \right) + k\left( {x - 2y + 3} \right) = 0,\,k \in R,$$ is a:

A

circle of radius $$\sqrt 2 $$.

B

circle of radius $$\sqrt 3 $$.

C

straight line parallel to $$x$$-axis

D

straight line parallel to $$y$$-axis

Intersection point of $$2x - 3y + 4 = 0$$

and $$x-2y+3=0$$ is $$(1, 2)$$

Since, $$P$$ is the fixed point for given family of lines

So, $$PB=PA$$

$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = {\left( {2 - 1} \right)^2} + {\left( {3 - 2} \right)^2}$$

$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = 1 + 1 = 2$$

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

$${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$$

Therefore, given locus is a circle with center $$(1, 2)$$ and radius $$\sqrt 2 .$$

and $$x-2y+3=0$$ is $$(1, 2)$$

Since, $$P$$ is the fixed point for given family of lines

So, $$PB=PA$$

$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = {\left( {2 - 1} \right)^2} + {\left( {3 - 2} \right)^2}$$

$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = 1 + 1 = 2$$

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

$${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$$

Therefore, given locus is a circle with center $$(1, 2)$$ and radius $$\sqrt 2 .$$

4

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)

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

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Complex Numbers

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Probability

Statistics

Mathematical Reasoning

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Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

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Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations