1
MCQ (Single Correct Answer)

### JEE Main 2016 (Online) 10th April Morning Slot

Equation of the tangent to the circle, at the point (1, −1), whose centre is the point of intersection of the straight lines x − y = 1 and 2x + y = 3 is :
A
4x + y − 3 = 0
B
x + 4y + 3 = 0
C
3x − y − 4 = 0
D
x − 3y − 4 = 0

## Explanation

Point of intersection of lines

x $-$ y = 1   and  2x + y = 3 is $\left( {{4 \over 3},{1 \over 3}} \right)$

Slope of OP = ${{{1 \over 3} + 1} \over {{4 \over 3} - 1}}$ = ${{{4 \over 3}} \over {{1 \over 3}}}$ = 4

Slope of tangent = $-$ ${1 \over 4}$

Equation of tangent    y + 1 = $-$ ${1 \over 4}$ (x $-$ 1)

4y + 4 = $-$ x + 1

x + 4y + 3 = 0

2
MCQ (Single Correct Answer)

### JEE Main 2017 (Offline)

The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y = |x| is :
A
$2\left( {\sqrt 2 - 1} \right)$
B
$4\left( {\sqrt 2 - 1} \right)$
C
$4\left( {\sqrt 2 + 1} \right)$
D
$2\left( {\sqrt 2 + 1} \right)$

## Explanation

Let the radius of circle with least area be r.

Then, the coordinate of the center = (0, b)

$\therefore$ The equation of circle be x2 + (y – b)2 = r2

Distance of perpendiculur from (0, 4) to y = x line = r

$\Rightarrow$ $\left| {{{ - b} \over {\sqrt 2 }}} \right| = r$

$\Rightarrow$ b = ${\sqrt 2 r}$

Circle passes through (0, 4),

$\therefore$ 0 + (4 – b)2 = r2

$\Rightarrow$ 4 - b = r

$\Rightarrow$ 4 - ${\sqrt 2 r}$ = r

$\Rightarrow$ r = ${4 \over {\sqrt 2 + 1}}$ = $4\left( {\sqrt 2 - 1} \right)$
3
MCQ (Single Correct Answer)

### 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$\sqrt 3$
4
MCQ (Single Correct Answer)

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

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

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

= ${{3 \over 4}}$

### EXAM MAP

#### Joint Entrance Examination

JEE Advanced JEE Main

#### Graduate Aptitude Test in Engineering

GATE CE GATE ECE GATE ME GATE IN GATE EE GATE CSE GATE PI

NEET