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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2011

MCQ (Single Correct Answer)
The two circles x2 + y2 = ax, and x2 + y2 = c2 (c > 0) touch each other if
A
| a | = c
B
a = 2c
C
| a | = 2c
D
2 | a | = c

Explanation

As center of one circle is $$\left( {0,0} \right)$$ and other circle passes through $$(0,0),$$ therefore

Also $${C_1}\left( {{a \over 2},0} \right){C_2}\left( {0,0} \right)$$

$${r_1} = {a \over 2}{r_2} = C$$

$${C_1}{C_2} = {r_1} - {r_2} = {a \over 2}$$

$$ \Rightarrow C - {a \over 2} = {a \over 2}$$

$$ \Rightarrow C = a$$

If the two circles touch each other, then they must touch each other internally.
2

AIEEE 2010

MCQ (Single Correct Answer)
The circle $${x^2} + {y^2} = 4x + 8y + 5$$ intersects the line $$3x - 4y - m$$ at two distinct points if
A
$$ - 35 < m < 15$$
B
$$ 15 < m < 65$$
C
$$ 35 < m < 85$$
D
$$ - 85 < m < -35$$

Explanation

Circle $${x^2} + {y^2} - 4x - 8y - 5 = 0$$

Center $$=(2,4),$$ Radius $$ = \sqrt {4 + 16 + 5} = 5$$

If circle is intersecting line $$3x-4y=m,$$ at two distinct points.

$$ \Rightarrow $$ length of perpendicular from center to the line $$ < $$ radius

$$ \Rightarrow {{\left| {6 - 16 - m} \right|} \over 5} < 5 \Rightarrow \left| {10 + m} \right| < 25$$

$$ \Rightarrow - 25 < m + 10 < 25 \Rightarrow - 35 < m < 15$$
3

AIEEE 2009

MCQ (Single Correct Answer)
If $$P$$ and $$Q$$ are the points of intersection of the circles
$${x^2} + {y^2} + 3x + 7y + 2p - 5 = 0$$ and $${x^2} + {y^2} + 2x + 2y - {p^2} = 0$$ then there is a circle passing through $$P,Q $$ and $$(1, 1)$$ for:
A
all except one value of $$p$$
B
all except two values of $$p$$
C
exactly one value of $$p$$
D
all values of $$p$$

Explanation

The given circles are

$${S_1} \equiv {x^2} + {y^2} + 3x + 7y + 2p - 5 = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$${S_2} \equiv {x^2} + {y^2} + 2x + 2y - {p^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

$$\therefore$$ Equation of common chord $$PQ$$ is $${S_1} - {S_2} = 0$$

$$ \Rightarrow L \equiv x + 5y + {p^2} + 2p - 5 = 0$$

$$ \Rightarrow $$ Equation of circle passing through $$P$$ and $$Q$$ is

$${S_1} + \lambda \,\,L = 0$$

$$ \Rightarrow \left( {{x^2} + {y^2} + 3x + 7y + 2p - 5} \right) + \lambda $$

$$\left( {x + 5y + {p^2} + 2p - 5} \right) = 0$$

As it passes through $$\left( {1,1} \right),$$ therefore

$$ \Rightarrow \left( {7 + 2p} \right) + \lambda \left( {2p + {p^2} + 1} \right) = 0$$

$$ \Rightarrow \lambda = - {{2p + 7} \over {\left( {p + 1} \right)}},$$

which does not exist for $$p=-1$$
4

AIEEE 2008

MCQ (Single Correct Answer)
The differential equation of the family of circles with fixed radius $$5$$ units and centre on the line $$y = 2$$ is
A
$$\left( {x - 2} \right){y^2} = 25 - {\left( {y - 2} \right)^2}$$
B
$$\left( {y - 2} \right){y^2} = 25 - {\left( {y - 2} \right)^2}$$
C
$${\left( {y - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}$$
D
$${\left( {x - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}$$

Explanation

Let the center of the circle be $$(h, 2)$$

$$\therefore$$ Equation of circle is

$${\left( {x - h} \right)^2} + \left( {y - 2} \right){}^2 = 25\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

Differentiating with respect to $$x,$$ we get

$$2\left( {x - h} \right) + 2\left( {y - 2} \right){{dy} \over {dx}} = 0$$

$$ \Rightarrow x - h = - \left( {y - 2} \right){{dy} \over {dx}}$$

Substituting in equation $$(1)$$ we get

$${\left( {y - 2} \right)^2}{\left( {{{dy} \over {dx}}} \right)^2} + {\left( {y - 2} \right)^2} = 25$$

$$ \Rightarrow {\left( {y - 2} \right)^2}{\left( {y'} \right)^2} = 25 - {\left( {y - 2} \right)^2}$$

Questions Asked from Circle

On those following papers in MCQ (Single Correct Answer)
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AIEEE 2007 (1)
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