1

### JEE Main 2019 (Online) 11th January Morning Slot

A square is inscribed in the circle x2 + y2 – 6x + 8y – 103 = 0 with its sides parallel to the coordinate axes. Then the distance of the vertex of this square which is nearest to the origin is :
A
$\sqrt {137}$
B
6
C
$\sqrt {41}$
D
13

## Explanation R $= \sqrt {9 + 16 + 103} = 8\sqrt 2$

OA $= 13$

OB $= \sqrt {265}$

OC $= \sqrt {137}$

OD $= \sqrt {41}$
2

### JEE Main 2019 (Online) 11th January Morning Slot

Two circles with equal radii are intersecting at the points (0, 1) and (0, –1). The tangent at the point (0, 1) to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is :
A
$2\sqrt 2$
B
$\sqrt 2$
C
2
D
1

## Explanation In $\Delta$APO

${\left( {{{\sqrt 2 r} \over 2}} \right)^2} + {1^2} = {r^2}$

$\Rightarrow$  $r = \sqrt 2$

So distance between centres $= \sqrt 2 r = 2$
3

### JEE Main 2019 (Online) 12th January Morning Slot

If a variable line, 3x + 4y – $\lambda$ = 0 is such that the two circles x2 + y2 – 2x – 2y + 1 = 0 and x2 + y2 – 18x – 2y + 78 = 0 are on its opposite sides, then the set of all values of $\lambda$ is the interval :
A
(23, 31)
B
(2, 17)
C
[13, 23]
D
[12, 21]

## Explanation

Centre of circles are opposite side of line

(3 + 4 $-$ $\lambda$) (27 + 4 $-$ $\lambda$) < 0

($\lambda$ $-$ 7) ($\lambda$ $-$ 31) < 0

$\lambda$ $\in$ (7, 31)

distance from S1

$\left| {{{3 + 4 - \lambda } \over 5}} \right| \ge 1 \Rightarrow \lambda \in ( - \infty ,2] \cup [(12,\infty ]$

distance from S2

$\left| {{{27 + 4 - \lambda } \over 5}} \right| \ge 2 \Rightarrow \lambda \in ( - \infty ,21] \cup [41,\infty )$

so  $\lambda \in \left[ {12,21} \right]$
4

### JEE Main 2019 (Online) 12th January Morning Slot

Let C1 and C2 be the centres of the circles x2 + y2 – 2x – 2y – 2 = 0 and x2 + y2 – 6x – 6y + 14 = 0 respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral PC1QC2 is:
A
4
B
6
C
9
D
8

## Explanation Area = 2 $\times$ ${1 \over 2}$.4 = 2