1
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

Equation of the tangent to the circle, at the point (1, −1), whose centre is the point of intersection of the straight lines x − y = 1 and 2x + y = 3 is :
A
4x + y − 3 = 0
B
x + 4y + 3 = 0
C
3x − y − 4 = 0
D
x − 3y − 4 = 0

Explanation

Point of intersection of lines

x $$-$$ y = 1   and  2x + y = 3 is $$\left( {{4 \over 3},{1 \over 3}} \right)$$

Slope of OP = $${{{1 \over 3} + 1} \over {{4 \over 3} - 1}}$$ = $${{{4 \over 3}} \over {{1 \over 3}}}$$ = 4

Slope of tangent = $$-$$ $${1 \over 4}$$

Equation of tangent    y + 1 = $$-$$ $${1 \over 4}$$ (x $$-$$ 1)

4y + 4 = $$-$$ x + 1

x + 4y + 3 = 0

2
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

The radius of a circle, having minimum area, which touches the curve y = 4 – x2 and the lines, y = |x| is :
A
$$2\left( {\sqrt 2 - 1} \right)$$
B
$$4\left( {\sqrt 2 - 1} \right)$$
C
$$4\left( {\sqrt 2 + 1} \right)$$
D
$$2\left( {\sqrt 2 + 1} \right)$$

Explanation

Let the radius of circle with least area be r.

Then, the coordinate of the center = (0, b)

$$ \therefore $$ The equation of circle be x2 + (y – b)2 = r2

Distance of perpendiculur from (0, 4) to y = x line = r

$$ \Rightarrow $$ $$\left| {{{ - b} \over {\sqrt 2 }}} \right| = r$$

$$ \Rightarrow $$ b = $${\sqrt 2 r}$$

Circle passes through (0, 4),

$$ \therefore $$ 0 + (4 – b)2 = r2

$$ \Rightarrow $$ 4 - b = r

$$ \Rightarrow $$ 4 - $${\sqrt 2 r}$$ = r

$$ \Rightarrow $$ r = $${4 \over {\sqrt 2 + 1}}$$ = $$4\left( {\sqrt 2 - 1} \right)$$
3
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 8th April Morning Slot

If a point P has co-ordinates (0, $$-$$2) and Q is any point on the circle, x2 + y2 $$-$$ 5x $$-$$ y + 5 = 0, then the maximum value of (PQ)2 is :
A
$${{25 + \sqrt 6 } \over 2}$$
B
14 + $$5\sqrt 3 $$
C
$${{47 + 10\sqrt 6 } \over 2}$$
D
8 + 5$$\sqrt 3 $$

Explanation

Given that x2 + y2 $$-$$ 5x $$-$$ y + 5 = 0

$$ \Rightarrow $$   (x $$-$$ 5/2)2 $$-$$ $${{25} \over 4}$$ + (y $$-$$ 1/2)2 $$-$$ 1/4 = 0

$$ \Rightarrow $$   (x $$-$$ 5/2)2 + (y $$-$$ 1/2)2 = 3/2

on circle [ Q $$ \equiv $$ (5/2 + $$\sqrt {3/2} $$ cos Q, $${1 \over 2}$$ + $$\sqrt {3/2} $$ sin Q)]

$$ \Rightarrow $$   PQ2 = $${\left( {{5 \over 2} + \sqrt {3/2} \cos Q} \right)^2}$$ + $${\left( {{5 \over 2} + \sqrt {3/2} \sin Q} \right)^2}$$

$$ \Rightarrow $$   PQ2 = $${{25} \over 2} + {3 \over 2} + 5\sqrt {3/2} $$ (cos Q + sinQ)

= 14 + 5$$\sqrt {3/2} $$ (cosQ + sinQ)

$$ \therefore $$   Maximum value of PQ2

= 14 + 5$$\sqrt {3/2} $$ $$ \times $$ $$\sqrt 2 $$ = 14 + 5$$\sqrt 3 $$
4
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 9th April Morning Slot

The equation
Im $$\left( {{{iz - 2} \over {z - i}}} \right)$$ + 1 = 0, z $$ \in $$ C, z $$ \ne $$ i
represents a part of a circle having radius equal to :
A
2
B
1
C
$${3 \over 4}$$
D
$${1 \over 2}$$

Explanation

Let z = x + iy

Then,

Im $$\left( {{{iz - 2} \over {z - i}}} \right)$$ + 1 = 0

$$ \Rightarrow $$ $${\mathop{\rm Im}\nolimits} \left[ {\left( {{{i\left( {x + iy} \right) - 2} \over {x + iy - i}}} \right)} \right] + 1 = 0$$

$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$$

$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{{ix - y - 2} \over {x + i\left( {y - 1} \right)}}} \right)\left( {{{x - i\left( {y - 1} \right)} \over {x - i\left( {y - 1} \right)}}} \right)} \right] + 1 = 0$$

$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ i{x^2} - {i^2}x\left( {y - 1} \right) - xy + \hfill \cr \,\,\,\,iy\left( {y - 1} \right) - 2x + i2\left( {y - 1} \right) \hfill \cr} \over {{x^2} - {i^2}{{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$

$$ \Rightarrow $$ $${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ i{x^2} + x\left( {y - 1} \right) - xy \hfill \cr \,\,\, - 2x + i\left( {y - 1} \right)\left( {y + 2} \right) \hfill \cr} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$

$$ \Rightarrow $$$${\mathop{\rm Im}\nolimits} \left[ {\left( {{\matrix{ x\left( {y - 1} \right) - xy\, - 2x \hfill \cr \,\, + i\left[ {\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \right] \hfill \cr} \over {{x^2} + {{\left( {y - 1} \right)}^2}}}} \right)} \right] + 1 = 0$$

$$ \Rightarrow $$$${{\left( {y - 1} \right)\left( {y + 2} \right) + {x^2}} \over {{x^2} + {{\left( {y - 1} \right)}^2}}} + 1 = 0$$

$$ \Rightarrow $$2x2 + 2y2 - y - 1 = 0

$$ \Rightarrow $$x2 + y2 - $$\left( {{1 \over 2}} \right)$$y - $$\left( {{1 \over 2}} \right)$$ = 0

$$ \therefore $$ Center of the circle is $$\left( {0,{1 \over 4}} \right)$$

$$ \therefore $$ Radius = $$\sqrt {{0^2} + {{\left( {{1 \over 4}} \right)}^2} + {1 \over 2}} $$

= $$\sqrt {{1 \over {16}} + {1 \over 2}} $$

= $$\sqrt {{9 \over {16}}} $$

= $${{3 \over 4}}$$

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