1

### JEE Main 2019 (Online) 10th January Evening Slot

If the area of an equilateral triangle inscribed in the circle x2 + y2 + 10x + 12y + c = 0 is $27\sqrt 3$ sq units then c is equal to
A
20
B
25
C
$-$ 25
D
13

## Explanation

$3\left( {{1 \over 2}{r^2}.\sin {{120}^o}} \right) = 27\sqrt 3$

${{{r^2}} \over 2}{{\sqrt 3 } \over 2} = {{27\sqrt 3 } \over 3}$

${r^2} = {{108} \over 3} = 36$

Radius   $= \sqrt {25 + 36 - C} = \sqrt {36}$

$C = 25$
2

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

The straight line x + 2y = 1 meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is
A
$4\sqrt 5$
B
${{\sqrt 5 } \over 2}$
C
$2\sqrt 5$
D
${{\sqrt 5 } \over 4}$

## Explanation

Equation of circle

(x $-$ 1) (x $-$ 0) + (y $-$ 0) (y $-$ ${1 \over 2}$) = 0

$\Rightarrow$  x2 + y2 $-$ x $-$ ${y \over 2}$ = 0

Equation of tangent of region is 2x + y = 0

$\ell$1 + $\ell$2 = ${2 \over {\sqrt 5 }} + {1 \over {2\sqrt 5 }}$

= ${{4 + 1} \over {2\sqrt 5 }} = {{\sqrt 5 } \over 2}$
3

### 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 $= \sqrt {41}$
4

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