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1

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 $$\sqrt 2 .$$

2

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)

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

3

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.

4

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 $$2h = {{10} \over 3}.$$

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Complex Numbers

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