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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2016 (Offline)

MCQ (Single Correct Answer)
If one of the diameters of the circle, given by the equation, $${x^2} + {y^2} - 4x + 6y - 12 = 0,$$ is a chord of a circle $$S$$, whose centre is at $$(-3, 2)$$, then the radius of $$S$$ is:
A
$$5$$
B
$$10$$
C
$$5\sqrt 2 $$
D
$$5\sqrt 3 $$

Explanation



Center of $$S$$ : $$O(-3, 2)$$ center of given circle $$A(2, -3)$$

$$ \Rightarrow OA = 5\sqrt 2 $$

Also $$AB=5$$ (as $$AB=r$$ of the given circle)

$$ \Rightarrow $$ Using pythagoras theorem in $$\Delta OAB$$

$$r = 5\sqrt 3 $$
2

JEE Main 2016 (Offline)

MCQ (Single Correct Answer)
The centres of those circles which touch the circle, $${x^2} + {y^2} - 8x - 8y - 4 = 0$$, externally and also touch the $$x$$-axis, lie on:
A
a hyperbola
B
a parabola
C
a circle
D
an ellipse which is not a circle

Explanation



For the given circle,

center : $$(4,4)$$

radius $$=6$$

$$6 + k = \sqrt {{{\left( {h - 4} \right)}^2} + {{\left( {k - 4} \right)}^2}} $$

$${\left( {h - 4} \right)^2} = 20k + 20$$

$$\therefore$$ locus of $$(h, k)$$ is

$${\left( {h - 4} \right)^2} = 20\left( {y + 1} \right),$$

which is parabola.
3

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
The number of common tangents to the circles $${x^2} + {y^2} - 4x - 6x - 12 = 0$$ and $${x^2} + {y^2} + 6x + 18y + 26 = 0,$$ is :
A
$$3$$
B
$$4$$
C
$$1$$
D
$$2$$

Explanation

$${x^2} + {y^2} - 4x - 6y - 12 = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

Center, $${c_1} = \left( {2,\,3} \right)$$ and Radius, $${r_1} = 5$$ units

$${x^2} + {y^2} + 6x + 18y + 26 = 0\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Center, $${c_2} = \left( { - 3, - 9} \right)$$ and Radius, $${r_2} = 8$$ units

$${C_1}{C_2} = \sqrt {{{\left( {2 + 3} \right)}^2} + {{\left( {3 + 9} \right)}^2}} = 13\,\,$$ units

$${r_1} + {r_2} = 5 + 8 = 13$$

$$\therefore$$ $${C_1}{C_2} = {r_1} + {r_2}$$



Therefore there are three common tangents.
4

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
Locus of the image of the point $$(2, 3)$$ in the line $$\left( {2x - 3y + 4} \right) + k\left( {x - 2y + 3} \right) = 0,\,k \in R,$$ is a:
A
circle of radius $$\sqrt 2 $$.
B
circle of radius $$\sqrt 3 $$.
C
straight line parallel to $$x$$-axis
D
straight line parallel to $$y$$-axis

Explanation

Intersection point of $$2x - 3y + 4 = 0$$

and $$x-2y+3=0$$ is $$(1, 2)$$



Since, $$P$$ is the fixed point for given family of lines

So, $$PB=PA$$

$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = {\left( {2 - 1} \right)^2} + {\left( {3 - 2} \right)^2}$$

$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = 1 + 1 = 2$$

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

$${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$$

Therefore, given locus is a circle with center $$(1, 2)$$ and radius $$\sqrt 2 .$$

Questions Asked from Circle

On those following papers in MCQ (Single Correct Answer)
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