1

### JEE Main 2019 (Online) 9th January Morning Slot

Three circles of radii a, b, c (a < b < c) touch each other externally. If they have x-axis as a common tangent, then :
A
a, b, c are in A.P.
B
$\sqrt a ,\sqrt b ,\sqrt c$ are in A.P
C
${1 \over {\sqrt b }} + {1 \over {\sqrt c }}$ = ${1 \over {\sqrt a }}$
D
${1 \over {\sqrt b }} = {1 \over {\sqrt a }} + {1 \over {\sqrt c }}$

## Explanation AB = AC + CB

$\sqrt {{{\left( {b + c} \right)}^2} - {{\left( {c - b} \right)}^2}}$ = $\sqrt {{{\left( {b + a} \right)}^2} - {{\left( {b - a} \right)}^2}}$
+ $\sqrt {{{\left( {c + a} \right)}^2} - {{\left( {c - a} \right)}^2}}$

$\Rightarrow$  $\sqrt {2bc}$ = $\sqrt {2ac}$ + $\sqrt {2ab}$

Dividing by $\sqrt {abc}$ we get.

$\Rightarrow$  ${1 \over {\sqrt a }}$ = ${1 \over {\sqrt b }}$ + ${1 \over {\sqrt c }}$
2

### JEE Main 2019 (Online) 9th January Evening Slot

If the circles

x2 + y2 $-$ 16x $-$ 20y + 164 = r2

and  (x $-$ 4)2 + (y $-$ 7)2 = 36

intersect at two distinct points, then :
A
r > 11
B
0 < r < 1
C
r = 11
D
1 < r < 11

## Explanation

Circles are x2 + y2 $-$ 16x $-$ 20y + 164 = r2 $\Rightarrow$ c1 (8, 10)

and (x $-$ 4)2 + (y $-$ 7)2 = 36

they intersect at two distinct points

$\left| {{r_1} - {r_2}} \right| < {c_1}{c_2} < {r_1} + {r_2}\left\{ {{c_1}{c_2} = \sqrt {16 + 9} = 5} \right\}$

Now  $\left| {r - 6} \right| < 5 < r + 6$

$\left| {r - 6} \right| < 5$

$\Rightarrow$  $- 5 < r - 6 < 5$

$\Rightarrow$  $1 < r < 11\,\,\,\,\,\,\,\,\,...(i)$

$5 < r + 6$

$- 1 < r\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

from (i) and (ii)

r $\in$ (1, 11)
3

### 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 = $\sqrt {16 + 25 - 16} = 5$
4

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

$C = 25$