1

### JEE Main 2017 (Online) 8th April Morning Slot

If a point P has co-ordinates (0, $-$2) and Q is any point on the circle, x2 + y2 $-$ 5x $-$ y + 5 = 0, then the maximum value of (PQ)2 is :
A
${{25 + \sqrt 6 } \over 2}$
B
14 + $5\sqrt 3$
C
${{47 + 10\sqrt 6 } \over 2}$
D
8 + 5$\sqrt 3$

## Explanation

Given that x2 + y2 $-$ 5x $-$ y + 5 = 0

$\Rightarrow$   (x $-$ 5/2)2 $-$ ${{25} \over 4}$ + (y $-$ 1/2)2 $-$ 1/4 = 0

$\Rightarrow$   (x $-$ 5/2)2 + (y $-$ 1/2)2 = 3/2

on circle [ Q $\equiv$ (5/2 + $\sqrt {3/2}$ cos Q, ${1 \over 2}$ + $\sqrt {3/2}$ sin Q)]

$\Rightarrow$   PQ2 = ${\left( {{5 \over 2} + \sqrt {3/2} \cos Q} \right)^2}$ + ${\left( {{5 \over 2} + \sqrt {3/2} \sin Q} \right)^2}$

$\Rightarrow$   PQ2 = ${{25} \over 2} + {3 \over 2} + 5\sqrt {3/2}$ (cos Q + sinQ)

= 14 + 5$\sqrt {3/2}$ (cosQ + sinQ)

$\therefore$   Maximum value of PQ2

= 14 + 5$\sqrt {3/2}$ $\times$ $\sqrt 2$ = 14 + 5$\sqrt 3$
2

### JEE Main 2017 (Online) 9th April Morning Slot

The equation
Im $\left( {{{iz - 2} \over {z - i}}} \right)$ + 1 = 0, z $\in$ C, z $\ne$ i
represents a part of a circle having radius equal to :
A
2
B
1
C
${3 \over 4}$
D
${1 \over 2}$

## Explanation

Let z = x + iy

Then,

Im $\left( {{{iz - 2} \over {z - i}}} \right)$ + 1 = 0

$\Rightarrow$ ${\mathop{\rm Im}\nolimits} \left[ {\left( {{{i\left( {x + iy} \right) - 2} \over {x + iy - i}}} \right)} \right] + 1 = 0$

$\Rightarrow $${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0 \Rightarrow$${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)\left( {{{x - i\left( {y - 1} \right)} \over {x - i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$

$\Rightarrow $${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ i{x^2} - {i^2}x\left( {y - 1} \right) - xy + \hfill \cr \,\,\,\,iy\left( {y - 1} \right) - 2x + i2\left( {y - 1} \right) \hfill \cr} \over {{x^2} - {i^2}{{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0 \Rightarrow {\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ i{x^2} + x\left( {y - 1} \right) - xy \hfill \cr \,\,\, - 2x + i\left( {y - 1} \right)\left( {y + 2} \right) \hfill \cr} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0 \Rightarrow$${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ x\left( {y - 1} \right) - xy\, - 2x \hfill \cr \,\, + i\left[ {\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \right] \hfill \cr} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$

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

$\Rightarrow$2x2 + 2y2 - y - 1 = 0

$\Rightarrow$x2 + y2 - $\left( {{1 \over 2}} \right)$y - $\left( {{1 \over 2}} \right)$ = 0

$\therefore$ Center of the circle is $\left( {0,{1 \over 4}} \right)$

$\therefore$ Radius = $\sqrt {{0^2} + {{\left( {{1 \over 4}} \right)}^2} + {1 \over 2}}$

= $\sqrt {{1 \over {16}} + {1 \over 2}}$

= $\sqrt {{9 \over {16}}}$

= ${{3 \over 4}}$
3

### JEE Main 2017 (Online) 9th April Morning Slot

A line drawn through the point P(4, 7) cuts the circle x2 + y2 = 9 at the points A and B. Then PA⋅PB is equal to :
A
53
B
56
C
74
D
65

## Explanation

P(4, 7).   Here, x = 4, y = 7

$\therefore$   PA $\times$ PB = PT2

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

$\Rightarrow$   PT = $\sqrt {16 + 49 - 9}$ = $\sqrt {56}$

$\Rightarrow$   PT2 = 56

$\therefore$   PA $\times$ PB = 56
4

### JEE Main 2017 (Online) 9th April Morning Slot

The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60o. If the area of the quadrilateral is $4\sqrt 3$, then the perimeter of the quadrilateral is :
A
12.5
B
13.2
C
12
D
13

## Explanation

Here; cos$\theta$ = ${{{a^2} + {b^2} - {c^2}} \over {2ab}}$

and  $\theta$ = 60o

$\Rightarrow$   cos 60o = ${{4 + 25 - {c^2}} \over {2.2.5}}$

$\Rightarrow$    10 = 29 $-$ c2

$\Rightarrow$   c2 = 19

$\Rightarrow$   c = $\sqrt {19}$

also;  cos$\theta$ = ${{{a^2} + {b^2} - {c^2}} \over {2ab}}$

and $\theta$ = 120o

$\Rightarrow$   $-$ ${1 \over 2}$ = ${{{a^2} + {b^2} - 19} \over {2ab}}$

$\Rightarrow$   a2 + b2 $-$ 19 = $-$ ab

$\Rightarrow$   a2 + b2 + ab = 19

$\therefore$   Area = ${1 \over 2} \times 2 \times 5$ sin 60 + ${1 \over 2}$ ab sin 120o = 4$\sqrt 3$

$\Rightarrow$   ${{5\sqrt 3 } \over 2} + {{ab\sqrt 3 } \over 4}$ = $4\sqrt 3$

$\Rightarrow$   ${{ab} \over 4}$ = 4 $-$ ${5 \over 2}$ = ${3 \over 2}$

$\Rightarrow$    ab = 6

$\therefore$   a2 + b2 = 13

$\Rightarrow$   a = 2, b = 3

Perimeter = Sum of all sides

= 2 + 5 + 2 + 3 = 12