1

### 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 $= \sqrt 2 r = 2$
2

### 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  $\lambda \in \left[ {12,21} \right]$
3

### 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
4

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

If a circle of radius R passes through the origin O and intersects the coordinates axes at A and B, then the locus of the foot of perpendicular from O on AB is :
A
(x2 + y2)2 = 4R2x2y2
B
(x2 + y2) (x + y) = R2xy
C
(x2 + y2)2 = 4Rx2y2
D
(x2 + y2)3 = 4R2x2y2

## Explanation Slope of AB = ${{ - h} \over k}$

Equation of AB is hx + ky = h2 + k2

A $\left( {{{{h^2} + {k^2}} \over h},0} \right),B\left( {0,{{{h^2} + {k^2}} \over k}} \right)$

AB = 2R

$\Rightarrow$   (h2 + k2)3 = 4R2h2k2

$\Rightarrow$  (x2 + y2)3 = 4R2x2y2