1

### JEE Main 2016 (Online) 9th April Morning Slot

A circle passes through (−2, 4) and touches the y-axis at (0, 2). Which one of the following equations can represent a diameter of this circle ?
A
4x + 5y − 6 = 0
B
2x − 3y + 10 = 0
C
3x + 4y − 3 = 0
D
5x + 2y + 4 = 0

## Explanation EF = perpendicular bisector of chord AB

BG = perpendicular to y-axis

Here C = center of the circle

mid-point of chord AB, D = ($-$ 1, 3)

slope of AB = ${{4 - 2} \over { - 2 - 0}}$ = $-$ 1

$\because$   EF  $\bot$ AB

$\therefore$   Slope of EF = 1

Equation of EF, y $-$ 3 = 1 (x + 1)

$\Rightarrow$    y = x + 4       . . . .(i)

Equation of BG

y = 2         . . . . (ii)

From equations (i) and (ii)

x = $-$ 2, y = 2

since C be the point of intersection of EF and BG, therefore center, C = ($-$ 2, 2)

Now coordinates of center C satiesfy the equation

2x $-$ 3y + 10 = 0

Hence 2x $-$ 3y + 10 = 0   is the equation of the diameter
2

### 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 3

### 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 = ${4 \over {\sqrt 2 + 1}}$ = $4\left( {\sqrt 2 - 1} \right)$
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### 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$\sqrt 3$