1

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

If a circle C passing through the point (4, 0) touches the circle x2 + y2 + 4x – 6y = 12 externally at the point (1, – 1), then the radius of C is -
A
5
B
2$\sqrt {5}$
C
4
D
$\sqrt {37}$

## Explanation

x2 + y2 + 4x $-$ 6y $-$ 12 = 0

Equation of tangent at (1, $-$ 1)

x $-$ y + 2(x + 1) $-$ 3(y $-$ 1) $-$ 12 = 0

3x $-$ 4y $-$ 7 = 0

$\therefore$   Equation of circle is

(x2 + y2 + 4x $-$ 6y $-$ 12) + $\lambda$ (3x $-$ 4y $-$ 7) = 0

It passes through (4, 0) :

(16 + 16 $-$ 12) + $\lambda$ (12 $-$ 7) = 0

$\Rightarrow$  20 + $\lambda$(5) = 0

$\Rightarrow$  $\lambda$ = $-$ 4

$\therefore$  (x2 + y2 + 4x $-$ 6y $-$ 12) $-$ 4(3x $-$ 4y $-$ 7) = 0

or   x2 + y2 $-$ 8x + 10y + 16 = 0

Radius = $\sqrt {16 + 25 - 16} = 5$
2

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

$C = 25$
3

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

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

### 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 $= \sqrt {41}$