1

### JEE Main 2017 (Online) 9th April Morning Slot

A line drawn through the point P(4, 7) cuts the circle x2 + y2 = 9 at the points A and B. Then PA⋅PB is equal to :
A
53
B
56
C
74
D
65

## Explanation

P(4, 7).   Here, x = 4, y = 7

$\therefore$   PA $\times$ PB = PT2

Also;   PT = $\sqrt {{x^2} + {y^2} - {{\left( {x - y} \right)}^2}}$

$\Rightarrow$   PT = $\sqrt {16 + 49 - 9}$ = $\sqrt {56}$

$\Rightarrow$   PT2 = 56

$\therefore$   PA $\times$ PB = 56
2

### JEE Main 2017 (Online) 9th April Morning Slot

The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60o. If the area of the quadrilateral is $4\sqrt 3$, then the perimeter of the quadrilateral is :
A
12.5
B
13.2
C
12
D
13

## Explanation Here; cos$\theta$ = ${{{a^2} + {b^2} - {c^2}} \over {2ab}}$

and  $\theta$ = 60o

$\Rightarrow$   cos 60o = ${{4 + 25 - {c^2}} \over {2.2.5}}$

$\Rightarrow$    10 = 29 $-$ c2

$\Rightarrow$   c2 = 19

$\Rightarrow$   c = $\sqrt {19}$

also;  cos$\theta$ = ${{{a^2} + {b^2} - {c^2}} \over {2ab}}$

and $\theta$ = 120o

$\Rightarrow$   $-$ ${1 \over 2}$ = ${{{a^2} + {b^2} - 19} \over {2ab}}$

$\Rightarrow$   a2 + b2 $-$ 19 = $-$ ab

$\Rightarrow$   a2 + b2 + ab = 19

$\therefore$   Area = ${1 \over 2} \times 2 \times 5$ sin 60 + ${1 \over 2}$ ab sin 120o = 4$\sqrt 3$

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

$\Rightarrow$   ${{ab} \over 4}$ = 4 $-$ ${5 \over 2}$ = ${3 \over 2}$

$\Rightarrow$    ab = 6

$\therefore$   a2 + b2 = 13

$\Rightarrow$   a = 2, b = 3

Perimeter = Sum of all sides

= 2 + 5 + 2 + 3 = 12
3

### JEE Main 2018 (Offline)

Let the orthocentre and centroid of a triangle be A(-3, 5) and B(3, 3) respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is :
A
${{3\sqrt 5 } \over 2}$
B
$\sqrt {10}$
C
$2\sqrt {10}$
D
$3\sqrt {{5 \over 2}}$

## Explanation

In a triangle, orthocentre, centroid and circumcenter are collinear and centroid divides orthocenter and circumcenter in 2 : 1 ratio. Here AB = 2 BC

$AB = \sqrt {{{\left( {3 + 3} \right)}^2} + {{\left( {3 - 5} \right)}^2}}$

$= \sqrt {36 + 4}$

$= \sqrt {40}$

$= 2\sqrt {10}$

$\therefore\,\,\,\,$ BC $= \sqrt {10}$

$\therefore\,\,\,\,$ AB + BC

$= 2\sqrt {10} + \sqrt {10}$

$= 3\sqrt {10}$

AC is the diameter of the circle.

$\therefore\,\,\,$ Radius $= {1 \over 2}\,\,$ AC

$= {1 \over 2} \times 3\sqrt {10}$

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

### JEE Main 2018 (Offline)

If the tangent at (1, 7) to the curve x2 = y - 6

touches the circle x2 + y2 + 16x + 12y + c = 0, then the value of c is :
A
95
B
195
C
185
D
85

## Explanation

NOTE :

Equation of tangent at (x1, y1) to the curve x2 = 4ay is

xx1 = 4a $\left( {{{y + {y_1}} \over 2}} \right)$

Now equation of tangent at (1, 7) to x2 = y $-$ 6 is

$x\,.\,1 = 4\,.\,{1 \over 4}\left( {{{y + 7} \over 2}} \right) - 6$

$\Rightarrow 2x = y + 7 - 12$

$\Rightarrow 2x - y + 5 = 0$

This tangent touches the circle. So, perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

For the circle,

${x^2} + {y^2} + 16x + 12y + C = 0$

center is ($-$8, $-$6)

and radius (r) = $\sqrt {{8^2} + {6^2} - c}$

$= \sqrt {100 - c}$

Distance of the tangent from the center of the circle

d = $\left| {{{2\left( { - 8} \right) - \left( { - 6} \right) + 5} \over {\sqrt {{2^2} + {1^2}} }}} \right|$

$= \left| {{{ - 16 + 6 + 5} \over {\sqrt 5 }}} \right|$

And we know d = r

$\therefore\,\,\,$ $\left| {{{ - 16 + 11} \over {\sqrt 5 }}} \right| = \sqrt {100 - c}$

$\Rightarrow \left| {{{ - 5} \over {\sqrt 5 }}} \right| = \sqrt {100 - c}$

$\Rightarrow \left| { - \sqrt 5 } \right| = \sqrt {100 - c}$

$\Rightarrow 5 = 100 - c$

$\Rightarrow c = 95$