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1

IIT-JEE 2008

MCQ (Single Correct Answer)
Consider $$\,{L_1}:\,\,2x\,\, + \,\,3y\, + \,p\,\, - \,\,3 = 0$$
$$\,{L_2}:\,\,2x\,\, + \,\,3y\, + \,p\,\, + \,\,3 = 0$$

where p is a real number, and $$\,C:\,{x^2}\, + \,{y^2}\, + \,6x\, - 10y\, + \,30 = 0$$
STATEMENT-1 : If line $${L_1}$$ is a chord of circle C, then line $${L_2}$$ is not always a diameter of circle C
and
STATEMENT-2 : If line $${L_1}$$ is a diameter of circle C, then line $${L_2}$$ is not a chord of circle C.

A
Statement-1 is True, Statement-2 is True; Statement-2 is a correct rexplanation for Statement-1
B
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct rexplanation for Statement-1
C
Statement-1 is True, Statement-2 is False
D
Statement-1 is False, Statement-2 is True

Explanation

Equation of circle C is

$${(x + 3)^2} + {(y - 5)^2} = 9 + 25 - 30 = 4$$

$$ \Rightarrow {(x + 3)^2} + {(y - 5)^2} = {2^2}$$

Centre = (3, $$-$$5)

If L1 is diameter, then $$2(3) + 3( - 5) + p - 3 = 0 \Rightarrow p = 12$$

$$\therefore$$ L1 is $$2x + 3y + 9 = 0$$

and L2 is $$2x + 3y + 15 = 0$$

Distance of centre of circle from $${L_2} = \left| {{{2(3) + 3( - 5) + 15} \over {\sqrt {{2^2} + {3^2}} }}} \right| = {6 \over {\sqrt {12} }} < 2$$ [radius of circle]

$$\therefore$$ L2 is a chord of circle C.

Statement 2 is false.

2

IIT-JEE 2008

MCQ (Single Correct Answer)
A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation $$\sqrt 3 x\, + \,y\, - \,6 = 0$$ and the point D is $$\left( {{{3\,\sqrt 3 } \over 2},\,{3 \over 2}} \right)$$. Further, it is given that the origin and the centre of C are on the same side of the line PQ.

Equations of the sides QR, RP are

A
$$y = {2 \over {\sqrt 3 }}\,x + \,1,\,\,y = \, - {2 \over {\sqrt 3 }}\,x - 1$$
B
$$y = {1 \over {\sqrt 3 }}\,x,\,\,y = \,0$$
C
$$y = {{\sqrt 3 } \over 2}\,x + \,1,\,\,y = \, - {{\sqrt 3 } \over 2}\,x - 1$$
D
$$y = \sqrt 3 \,x,\,\,y = \,0$$
3

IIT-JEE 2008

MCQ (Single Correct Answer)
A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation $$\sqrt 3 x\, + \,y\, - \,6 = 0$$ and the point D is $$\left( {{{3\,\sqrt 3 } \over 2},\,{3 \over 2}} \right)$$. Further, it is given that the origin and the centre of C are on the same side of the line PQ.

Ponits E and F are given by

A
$$\left( {{{\,\sqrt 3 } \over 2},\,{3 \over 2}} \right),\,\left( {\sqrt 3 ,\,0} \right)$$
B
$$\left( {{{\,\sqrt 3 } \over 2},\,{1 \over 2}} \right),\,\left( {\sqrt 3 ,\,0} \right)$$
C
$$\left( {{{\,\sqrt 3 } \over 2},\,{3 \over 2}} \right),\,\left( {{{\,\sqrt 3 } \over 2},\,{1 \over 2}} \right)$$
D
$$\left( {{{\,3} \over 2},\,{{\sqrt 3 } \over 2}} \right),\,\left( {{{\,\sqrt 3 } \over 2},\,{1 \over 2}} \right)$$
4

IIT-JEE 2008

MCQ (Single Correct Answer)
A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation $$\sqrt 3 x\, + \,y\, - \,6 = 0$$ and the point D is $$\left( {{{3\,\sqrt 3 } \over 2},\,{3 \over 2}} \right)$$. Further, it is given that the origin and the centre of C are on the same side of the line PQ.

The equation of circle C is

A
$${\left( {x\, - 2\sqrt 3 \,} \right)^2} + {(y - 1)^2} = 1$$
B
$${\left( {x\, - 2\sqrt 3 \,} \right)^2} + {(y + {1 \over 2})^2} = 1$$
C
$${\left( {x\, - \sqrt 3 \,} \right)^2} + {(y + 1)^2} = 1$$
D
$${\left( {x\, - \sqrt 3 \,} \right)^2} + {(y - 1)^2} = 1$$

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